GCSE Maths · Quadratic and Simultaneous Equations · Higher · Lesson 14.4

GCSE Maths: Problems leading to quadratic equations (Higher)

Part of the GCSE Maths quadratic and simultaneous equations chapter (Higher). Chapter overview · Foundation version of this lesson · whole chapter on one page · Download worksheet (PDF) · Answers (PDF)

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14.4

Problems leading to quadratic equations

Edexcel 1MA1: A18, A21
Grade 4-7

Quadratic equations appear whenever two lengths are multiplied together, for example in areas, or whenever two numbers are multiplied together, for example “consecutive numbers”. Here is a strategy that works for every problem in this section.

Choose a letter for the unknown and write down what it stands for, with units.
Write every other quantity in terms of that letter (a rectangle that is 3 cm longer than it is wide has length x + 3 if the width is x).
Form an equation from the information in the question (for instance “area = 54”).
Expand, rearrange to equal 0, factorise and solve exactly as in the last two sections.
Decide which answers make sense. A length cannot be negative, so a negative solution is rejected. Then answer the question that was asked.
Check in the words of the problem, not just in the equation.

Consecutive numbers follow each other with no gaps: 8, 9, 10. If the first is n the next is n + 1. Consecutive odd numbers (or consecutive even numbers) are 2 apart: n and n + 2. The product of two consecutive numbers is n(n + 1).

A rectangle that is 3 cm longer than it is wide. If the area is 54 cm², then x(x + 3) = 54.

Higher. Problems can also use Pythagoras’ theorem. For a right-angled triangle with shorter sides a and b and hypotenuse c, a² + b² = c² (see Chapter 20). When the sides are given as expressions with x, you must expand squared brackets such as (x + 7)² = (x + 7)(x + 7) carefully. It does not equal x² + 49.

Key ideas

Choose a letter, write the other quantities using it, form an equation, solve, then interpret.
Reject answers that cannot happen (a negative length or a negative number of items).
Consecutive whole numbers: n, n + 1. Consecutive odd (or even) numbers: n, n + 2.
Area of a rectangle = length × width; area of a triangle = ½ × base × height.
Check your answer against the words of the question.

Worked example 14.4.1

A rectangle has width x cm and length (x + 3) cm. Its area is 54 cm². Find the width and the length.

Key words. A quadratic equation is an equation whose highest power of the letter is 2 (it has an x² term), like x² + 5x + 6 = 0. To solve it means to find every value of x that makes it true. These values are called the solutions or roots. To factorise means to write it as brackets multiplied together.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
Let the width be x cm. Then the length is (x + 3) cm.
Form an equation from the information. Area of a rectangle = length × width.
x(x + 3) = 54
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand x(x + 3):
x × x = x²
x × 3 = 3x
so x(x + 3) = x² + 3x
The equation is now: x² + 3x = 54
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 54 from both sides: x² + 3x − 54 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 3 and c = -54.
We need two numbers that multiply to -54 (the last number) and add to 3 (the middle number). List the pairs that multiply to -54:
1 × -54 = -54, add: 1 + -54 = -53 ✗
-1 × 54 = -54, add: -1 + 54 = 53 ✗
2 × -27 = -54, add: 2 + -27 = -25 ✗
-2 × 27 = -54, add: -2 + 27 = 25 ✗
3 × -18 = -54, add: 3 + -18 = -15 ✗
-3 × 18 = -54, add: -3 + 18 = 15 ✗
6 × -9 = -54, add: 6 + -9 = -3 ✗
-6 × 9 = -54, add: -6 + 9 = 3 ✓ this pair works
Put the pair into two brackets: (x − 6)(x + 9).
Check by expanding: x² + 9x − 6x − 54 = x² + 3x − 54. ✓
x² + 3x − 54 = (x − 6)(x + 9)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x − 6)(x + 9) = 0
So (x − 6) = 0 or (x + 9) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 6 = 0
Add 6 to both sides: x = 6

x + 9 = 0
Subtract 9 from both sides: x = −9
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −9 or x = 6
Decide which solution makes sense. x is a width, so it cannot be negative. Reject x = −9.
x = 6 cm
Find the length and answer the question. The length is x + 3.
Length = 6 + 3 = 9 cm
Width = 6 cm, length = 9 cm

Check. Check in the words: width 6 cm and length 9 cm differ by 3 cm and the area is 6 × 9 = 54 cm². ✓

Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).

Worked example 14.4.2

The product of two consecutive positive whole numbers is 156. Find the numbers.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
Let the smaller number be n. The next whole number is n + 1.
Form an equation from the information. The product of the two numbers means they are multiplied.
n(n + 1) = 156
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand n(n + 1):
n × n = n²
n × 1 = n
so n(n + 1) = n² + n
The equation is now: n² + n = 156
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 156 from both sides: n² + n − 156 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form n² + bn + c, with b = 1 and c = -156.
We need two numbers that multiply to -156 (the last number) and add to 1 (the middle number). List the pairs that multiply to -156:
1 × -156 = -156, add: 1 + -156 = -155 ✗
-1 × 156 = -156, add: -1 + 156 = 155 ✗
2 × -78 = -156, add: 2 + -78 = -76 ✗
-2 × 78 = -156, add: -2 + 78 = 76 ✗
3 × -52 = -156, add: 3 + -52 = -49 ✗
-3 × 52 = -156, add: -3 + 52 = 49 ✗
4 × -39 = -156, add: 4 + -39 = -35 ✗
-4 × 39 = -156, add: -4 + 39 = 35 ✗
6 × -26 = -156, add: 6 + -26 = -20 ✗
-6 × 26 = -156, add: -6 + 26 = 20 ✗
12 × -13 = -156, add: 12 + -13 = -1 ✗
-12 × 13 = -156, add: -12 + 13 = 1 ✓ this pair works
Put the pair into two brackets: (n − 12)(n + 13).
Check by expanding: n² + 13n − 12n − 156 = n² + n − 156. ✓
n² + n − 156 = (n − 12)(n + 13)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(n − 12)(n + 13) = 0
So (n − 12) = 0 or (n + 13) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
n − 12 = 0
Add 12 to both sides: n = 12

n + 13 = 0
Subtract 13 from both sides: n = −13
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
n = −13 or n = 12
Decide which solution makes sense. The numbers are positive, so reject n = −13.
n = 12
Find both numbers. The numbers are n and n + 1.
12 and 12 + 1 = 13
The numbers are 12 and 13.

Check. Check: 12 × 13 = 156. ✓

Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).

Worked example 14.4.3

A triangle has base (x + 4) cm and height x cm. Its area is 30 cm². Find the height and the base.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
Let the height of the triangle be x cm. Then the base is (x + 4) cm.
Form an equation from the information. The area of a triangle is ½ × base × height. Here ½ × (x + 4) × x = 30. Multiply both sides by 2 to remove the half, so (x + 4) × x = 60.
x(x + 4) = 60
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand x(x + 4):
x × x = x²
x × 4 = 4x
so x(x + 4) = x² + 4x
The equation is now: x² + 4x = 60
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 60 from both sides: x² + 4x − 60 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 4 and c = -60.
We need two numbers that multiply to -60 (the last number) and add to 4 (the middle number). List the pairs that multiply to -60:
1 × -60 = -60, add: 1 + -60 = -59 ✗
-1 × 60 = -60, add: -1 + 60 = 59 ✗
2 × -30 = -60, add: 2 + -30 = -28 ✗
-2 × 30 = -60, add: -2 + 30 = 28 ✗
3 × -20 = -60, add: 3 + -20 = -17 ✗
-3 × 20 = -60, add: -3 + 20 = 17 ✗
4 × -15 = -60, add: 4 + -15 = -11 ✗
-4 × 15 = -60, add: -4 + 15 = 11 ✗
5 × -12 = -60, add: 5 + -12 = -7 ✗
-5 × 12 = -60, add: -5 + 12 = 7 ✗
6 × -10 = -60, add: 6 + -10 = -4 ✗
-6 × 10 = -60, add: -6 + 10 = 4 ✓ this pair works
Put the pair into two brackets: (x − 6)(x + 10).
Check by expanding: x² + 10x − 6x − 60 = x² + 4x − 60. ✓
x² + 4x − 60 = (x − 6)(x + 10)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x − 6)(x + 10) = 0
So (x − 6) = 0 or (x + 10) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 6 = 0
Add 6 to both sides: x = 6

x + 10 = 0
Subtract 10 from both sides: x = −10
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −10 or x = 6
Decide which solution makes sense. A height cannot be negative, so reject x = −10.
x = 6 cm
Find the base and answer. The base is x + 4.
Base = 10 cm
Height = 6 cm, base = 10 cm

Check. Area = ½ × 10 × 6 = 30 cm². ✓

Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).

Worked example 14.4.4

The product of two consecutive positive odd numbers is 143. Find the numbers.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
Let the smaller number be n. The next consecutive odd number is n + 2 (odd numbers are 2 apart).
Form an equation from the information. The product of the two numbers means they are multiplied.
n(n + 2) = 143
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand n(n + 2):
n × n = n²
n × 2 = 2n
so n(n + 2) = n² + 2n
The equation is now: n² + 2n = 143
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 143 from both sides: n² + 2n − 143 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form n² + bn + c, with b = 2 and c = -143.
We need two numbers that multiply to -143 (the last number) and add to 2 (the middle number). List the pairs that multiply to -143:
1 × -143 = -143, add: 1 + -143 = -142 ✗
-1 × 143 = -143, add: -1 + 143 = 142 ✗
11 × -13 = -143, add: 11 + -13 = -2 ✗
-11 × 13 = -143, add: -11 + 13 = 2 ✓ this pair works
Put the pair into two brackets: (n − 11)(n + 13).
Check by expanding: n² + 13n − 11n − 143 = n² + 2n − 143. ✓
n² + 2n − 143 = (n − 11)(n + 13)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(n − 11)(n + 13) = 0
So (n − 11) = 0 or (n + 13) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
n − 11 = 0
Add 11 to both sides: n = 11

n + 13 = 0
Subtract 13 from both sides: n = −13
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
n = −13 or n = 11
Decide which solution makes sense. The numbers are positive, so reject n = −13.
n = 11
Find both numbers. The numbers are n and n + 2.
11 and 11 + 2 = 13
The numbers are 11 and 13.

Check. Check: 11 × 13 = 143. ✓

Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).

Worked example 14.4.5

The sides of a right-angled triangle are x cm, (x + 1) cm and (x + 2) cm, where (x + 2) cm is the longest side. Find x.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
Let the shortest side be x cm. The other two sides are (x + 1) cm and the hypotenuse (x + 2) cm.
Form an equation from the information. Pythagoras: (shorter side)² + (other side)² = (hypotenuse)². The longest side, the hypotenuse, is (x + 2).
x² + (x + 1)² = (x + 2)²
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand (x + 1)(x + 1):
First: x × x = x²
Outer: x × 1 = x
Inner: 1 × x = x
Last: 1 × 1 = 1
so (x + 1)(x + 1) = x² + 2x + 1

Expand (x + 2)(x + 2):
First: x × x = x²
Outer: x × 2 = 2x
Inner: 2 × x = 2x
Last: 2 × 2 = 4
so (x + 2)(x + 2) = x² + 4x + 4
The equation is now: 2x² + 2x + 1 = x² + 4x + 4
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract x² from both sides: 2x² + 2x + 1 − x² = 4x + 4
Subtract 4x from both sides: 2x² + 2x + 1 − x² − 4x = 4
Subtract 4 from both sides: 2x² + 2x + 1 − x² − 4x − 4 = 0
Collect like terms: x² − 2x − 3 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = -2 and c = -3.
We need two numbers that multiply to -3 (the last number) and add to -2 (the middle number). List the pairs that multiply to -3:
1 × -3 = -3, add: 1 + -3 = -2 ✓ this pair works
Put the pair into two brackets: (x + 1)(x − 3).
Check by expanding: x² − 3x + 1x − 3 = x² − 2x − 3. ✓
x² − 2x − 3 = (x + 1)(x − 3)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x + 1)(x − 3) = 0
So (x − 3) = 0 or (x + 1) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 3 = 0
Add 3 to both sides: x = 3

x + 1 = 0
Subtract 1 from both sides: x = −1
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −1 or x = 3
Decide which solution makes sense. A length cannot be negative, so reject x = −1.
x = 3 cm
Find all three sides. Put x = 3 into each expression.
Sides: 3 cm, 3 + 1 = 4 cm and 3 + 2 = 5 cm
The sides are 3 cm, 4 cm and 5 cm.

Check. 3² + 4² = 9 + 16 = 25 = 5². ✓

Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).

Watch out

Always reject impossible answers, and always answer the question that was asked. If x is the width, the question may want the length or the area, so do not stop at x. Also remember (x + 1)² is not x² + 1: it is x² + 2x + 1.

Exercise 14D Problems leading to quadratic equations

Non-calculator

PSSolve these number problems. In each case form an equation first.

(a)The product of two consecutive positive whole numbers is 72. Find the numbers.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
Let the smaller number be n. The next whole number is n + 1.
Form an equation from the information. The product of the two numbers means they are multiplied.
n(n + 1) = 72
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand n(n + 1):
n × n = n²
n × 1 = n
so n(n + 1) = n² + n
The equation is now: n² + n = 72
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 72 from both sides: n² + n − 72 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form n² + bn + c, with b = 1 and c = -72.
We need two numbers that multiply to -72 (the last number) and add to 1 (the middle number). List the pairs that multiply to -72:
1 × -72 = -72, add: 1 + -72 = -71 ✗
-1 × 72 = -72, add: -1 + 72 = 71 ✗
2 × -36 = -72, add: 2 + -36 = -34 ✗
-2 × 36 = -72, add: -2 + 36 = 34 ✗
3 × -24 = -72, add: 3 + -24 = -21 ✗
-3 × 24 = -72, add: -3 + 24 = 21 ✗
4 × -18 = -72, add: 4 + -18 = -14 ✗
-4 × 18 = -72, add: -4 + 18 = 14 ✗
6 × -12 = -72, add: 6 + -12 = -6 ✗
-6 × 12 = -72, add: -6 + 12 = 6 ✗
8 × -9 = -72, add: 8 + -9 = -1 ✗
-8 × 9 = -72, add: -8 + 9 = 1 ✓ this pair works
Put the pair into two brackets: (n − 8)(n + 9).
Check by expanding: n² + 9n − 8n − 72 = n² + n − 72. ✓
n² + n − 72 = (n − 8)(n + 9)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(n − 8)(n + 9) = 0
So (n − 8) = 0 or (n + 9) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
n − 8 = 0
Add 8 to both sides: n = 8

n + 9 = 0
Subtract 9 from both sides: n = −9
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
n = −9 or n = 8
Decide which solution makes sense. The numbers are positive, so reject n = −9.
n = 8
Find both numbers. The numbers are n and n + 1.
8 and 8 + 1 = 9
The numbers are 8 and 9.

Check. Check: 8 × 9 = 72. ✓

Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).

(b)The product of two consecutive positive whole numbers is 110. Find the numbers.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
Let the smaller number be n. The next whole number is n + 1.
Form an equation from the information. The product of the two numbers means they are multiplied.
n(n + 1) = 110
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand n(n + 1):
n × n = n²
n × 1 = n
so n(n + 1) = n² + n
The equation is now: n² + n = 110
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 110 from both sides: n² + n − 110 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form n² + bn + c, with b = 1 and c = -110.
We need two numbers that multiply to -110 (the last number) and add to 1 (the middle number). List the pairs that multiply to -110:
1 × -110 = -110, add: 1 + -110 = -109 ✗
-1 × 110 = -110, add: -1 + 110 = 109 ✗
2 × -55 = -110, add: 2 + -55 = -53 ✗
-2 × 55 = -110, add: -2 + 55 = 53 ✗
5 × -22 = -110, add: 5 + -22 = -17 ✗
-5 × 22 = -110, add: -5 + 22 = 17 ✗
10 × -11 = -110, add: 10 + -11 = -1 ✗
-10 × 11 = -110, add: -10 + 11 = 1 ✓ this pair works
Put the pair into two brackets: (n − 10)(n + 11).
Check by expanding: n² + 11n − 10n − 110 = n² + n − 110. ✓
n² + n − 110 = (n − 10)(n + 11)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(n − 10)(n + 11) = 0
So (n − 10) = 0 or (n + 11) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
n − 10 = 0
Add 10 to both sides: n = 10

n + 11 = 0
Subtract 11 from both sides: n = −11
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
n = −11 or n = 10
Decide which solution makes sense. The numbers are positive, so reject n = −11.
n = 10
Find both numbers. The numbers are n and n + 1.
10 and 10 + 1 = 11
The numbers are 10 and 11.

Check. Check: 10 × 11 = 110. ✓

Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).

(c)The product of two consecutive positive whole numbers is 210. Find the numbers.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
Let the smaller number be n. The next whole number is n + 1.
Form an equation from the information. The product of the two numbers means they are multiplied.
n(n + 1) = 210
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand n(n + 1):
n × n = n²
n × 1 = n
so n(n + 1) = n² + n
The equation is now: n² + n = 210
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 210 from both sides: n² + n − 210 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form n² + bn + c, with b = 1 and c = -210.
We need two numbers that multiply to -210 (the last number) and add to 1 (the middle number). List the pairs that multiply to -210:
1 × -210 = -210, add: 1 + -210 = -209 ✗
-1 × 210 = -210, add: -1 + 210 = 209 ✗
2 × -105 = -210, add: 2 + -105 = -103 ✗
-2 × 105 = -210, add: -2 + 105 = 103 ✗
3 × -70 = -210, add: 3 + -70 = -67 ✗
-3 × 70 = -210, add: -3 + 70 = 67 ✗
5 × -42 = -210, add: 5 + -42 = -37 ✗
-5 × 42 = -210, add: -5 + 42 = 37 ✗
6 × -35 = -210, add: 6 + -35 = -29 ✗
-6 × 35 = -210, add: -6 + 35 = 29 ✗
7 × -30 = -210, add: 7 + -30 = -23 ✗
-7 × 30 = -210, add: -7 + 30 = 23 ✗
10 × -21 = -210, add: 10 + -21 = -11 ✗
-10 × 21 = -210, add: -10 + 21 = 11 ✗
14 × -15 = -210, add: 14 + -15 = -1 ✗
-14 × 15 = -210, add: -14 + 15 = 1 ✓ this pair works
Put the pair into two brackets: (n − 14)(n + 15).
Check by expanding: n² + 15n − 14n − 210 = n² + n − 210. ✓
n² + n − 210 = (n − 14)(n + 15)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(n − 14)(n + 15) = 0
So (n − 14) = 0 or (n + 15) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
n − 14 = 0
Add 14 to both sides: n = 14

n + 15 = 0
Subtract 15 from both sides: n = −15
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
n = −15 or n = 14
Decide which solution makes sense. The numbers are positive, so reject n = −15.
n = 14
Find both numbers. The numbers are n and n + 1.
14 and 14 + 1 = 15
The numbers are 14 and 15.

Check. Check: 14 × 15 = 210. ✓

Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).

Done

PSSolve these rectangle problems.

(a)A rectangle has width x cm and length (x + 2) cm. Its area is 24 cm². Find the width and the length.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
Let the width be x cm. Then the length is (x + 2) cm.
Form an equation from the information. Area of a rectangle = length × width.
x(x + 2) = 24
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand x(x + 2):
x × x = x²
x × 2 = 2x
so x(x + 2) = x² + 2x
The equation is now: x² + 2x = 24
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 24 from both sides: x² + 2x − 24 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 2 and c = -24.
We need two numbers that multiply to -24 (the last number) and add to 2 (the middle number). List the pairs that multiply to -24:
1 × -24 = -24, add: 1 + -24 = -23 ✗
-1 × 24 = -24, add: -1 + 24 = 23 ✗
2 × -12 = -24, add: 2 + -12 = -10 ✗
-2 × 12 = -24, add: -2 + 12 = 10 ✗
3 × -8 = -24, add: 3 + -8 = -5 ✗
-3 × 8 = -24, add: -3 + 8 = 5 ✗
4 × -6 = -24, add: 4 + -6 = -2 ✗
-4 × 6 = -24, add: -4 + 6 = 2 ✓ this pair works
Put the pair into two brackets: (x − 4)(x + 6).
Check by expanding: x² + 6x − 4x − 24 = x² + 2x − 24. ✓
x² + 2x − 24 = (x − 4)(x + 6)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x − 4)(x + 6) = 0
So (x − 4) = 0 or (x + 6) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 4 = 0
Add 4 to both sides: x = 4

x + 6 = 0
Subtract 6 from both sides: x = −6
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −6 or x = 4
Decide which solution makes sense. x is a width, so it cannot be negative. Reject x = −6.
x = 4 cm
Find the length and answer the question. The length is x + 2.
Length = 4 + 2 = 6 cm
Width = 4 cm, length = 6 cm

Check. Check in the words: width 4 cm and length 6 cm differ by 2 cm and the area is 4 × 6 = 24 cm². ✓

Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).

(b)A rectangle has width x cm and length (x + 5) cm. Its area is 36 cm². Find the width and the length.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
Let the width be x cm. Then the length is (x + 5) cm.
Form an equation from the information. Area of a rectangle = length × width.
x(x + 5) = 36
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand x(x + 5):
x × x = x²
x × 5 = 5x
so x(x + 5) = x² + 5x
The equation is now: x² + 5x = 36
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 36 from both sides: x² + 5x − 36 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 5 and c = -36.
We need two numbers that multiply to -36 (the last number) and add to 5 (the middle number). List the pairs that multiply to -36:
1 × -36 = -36, add: 1 + -36 = -35 ✗
-1 × 36 = -36, add: -1 + 36 = 35 ✗
2 × -18 = -36, add: 2 + -18 = -16 ✗
-2 × 18 = -36, add: -2 + 18 = 16 ✗
3 × -12 = -36, add: 3 + -12 = -9 ✗
-3 × 12 = -36, add: -3 + 12 = 9 ✗
4 × -9 = -36, add: 4 + -9 = -5 ✗
-4 × 9 = -36, add: -4 + 9 = 5 ✓ this pair works
Put the pair into two brackets: (x − 4)(x + 9).
Check by expanding: x² + 9x − 4x − 36 = x² + 5x − 36. ✓
x² + 5x − 36 = (x − 4)(x + 9)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x − 4)(x + 9) = 0
So (x − 4) = 0 or (x + 9) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 4 = 0
Add 4 to both sides: x = 4

x + 9 = 0
Subtract 9 from both sides: x = −9
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −9 or x = 4
Decide which solution makes sense. x is a width, so it cannot be negative. Reject x = −9.
x = 4 cm
Find the length and answer the question. The length is x + 5.
Length = 4 + 5 = 9 cm
Width = 4 cm, length = 9 cm

Check. Check in the words: width 4 cm and length 9 cm differ by 5 cm and the area is 4 × 9 = 36 cm². ✓

Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).

(c)A rectangle has width x cm and length (x + 3) cm. Its area is 40 cm². Find the width and the length.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
Let the width be x cm. Then the length is (x + 3) cm.
Form an equation from the information. Area of a rectangle = length × width.
x(x + 3) = 40
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand x(x + 3):
x × x = x²
x × 3 = 3x
so x(x + 3) = x² + 3x
The equation is now: x² + 3x = 40
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 40 from both sides: x² + 3x − 40 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 3 and c = -40.
We need two numbers that multiply to -40 (the last number) and add to 3 (the middle number). List the pairs that multiply to -40:
1 × -40 = -40, add: 1 + -40 = -39 ✗
-1 × 40 = -40, add: -1 + 40 = 39 ✗
2 × -20 = -40, add: 2 + -20 = -18 ✗
-2 × 20 = -40, add: -2 + 20 = 18 ✗
4 × -10 = -40, add: 4 + -10 = -6 ✗
-4 × 10 = -40, add: -4 + 10 = 6 ✗
5 × -8 = -40, add: 5 + -8 = -3 ✗
-5 × 8 = -40, add: -5 + 8 = 3 ✓ this pair works
Put the pair into two brackets: (x − 5)(x + 8).
Check by expanding: x² + 8x − 5x − 40 = x² + 3x − 40. ✓
x² + 3x − 40 = (x − 5)(x + 8)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x − 5)(x + 8) = 0
So (x − 5) = 0 or (x + 8) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 5 = 0
Add 5 to both sides: x = 5

x + 8 = 0
Subtract 8 from both sides: x = −8
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −8 or x = 5
Decide which solution makes sense. x is a width, so it cannot be negative. Reject x = −8.
x = 5 cm
Find the length and answer the question. The length is x + 3.
Length = 5 + 3 = 8 cm
Width = 5 cm, length = 8 cm

Check. Check in the words: width 5 cm and length 8 cm differ by 3 cm and the area is 5 × 8 = 40 cm². ✓

Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).

(d)A rectangle has width x cm and length (x + 1) cm. Its area is 56 cm². Find the width and the length.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
Let the width be x cm. Then the length is (x + 1) cm.
Form an equation from the information. Area of a rectangle = length × width.
x(x + 1) = 56
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand x(x + 1):
x × x = x²
x × 1 = x
so x(x + 1) = x² + x
The equation is now: x² + x = 56
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 56 from both sides: x² + x − 56 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 1 and c = -56.
We need two numbers that multiply to -56 (the last number) and add to 1 (the middle number). List the pairs that multiply to -56:
1 × -56 = -56, add: 1 + -56 = -55 ✗
-1 × 56 = -56, add: -1 + 56 = 55 ✗
2 × -28 = -56, add: 2 + -28 = -26 ✗
-2 × 28 = -56, add: -2 + 28 = 26 ✗
4 × -14 = -56, add: 4 + -14 = -10 ✗
-4 × 14 = -56, add: -4 + 14 = 10 ✗
7 × -8 = -56, add: 7 + -8 = -1 ✗
-7 × 8 = -56, add: -7 + 8 = 1 ✓ this pair works
Put the pair into two brackets: (x − 7)(x + 8).
Check by expanding: x² + 8x − 7x − 56 = x² + x − 56. ✓
x² + x − 56 = (x − 7)(x + 8)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x − 7)(x + 8) = 0
So (x − 7) = 0 or (x + 8) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 7 = 0
Add 7 to both sides: x = 7

x + 8 = 0
Subtract 8 from both sides: x = −8
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −8 or x = 7
Decide which solution makes sense. x is a width, so it cannot be negative. Reject x = −8.
x = 7 cm
Find the length and answer the question. The length is x + 1.
Length = 7 + 1 = 8 cm
Width = 7 cm, length = 8 cm

Check. Check in the words: width 7 cm and length 8 cm differ by 1 cm and the area is 7 × 8 = 56 cm². ✓

Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).

Done

PSThe square of a number is 5 more than four times the number. Find the possible values of the number.

Method. To solve a quadratic equation by factorising: (1) get one side equal to 0, (2) factorise, (3) use the null factor law to make each bracket equal to 0, (4) solve the little equations.

Write the sentence as an equation. “The square of a number” is x². “5 more than 4 times the number” is 4x + 5.
x² = 4x + 5
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 4x from both sides: x² − 4x = 5
Subtract 5 from both sides: x² − 4x − 5 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = -4 and c = -5.
We need two numbers that multiply to -5 (the last number) and add to -4 (the middle number). List the pairs that multiply to -5:
1 × -5 = -5, add: 1 + -5 = -4 ✓ this pair works
Put the pair into two brackets: (x + 1)(x − 5).
Check by expanding: x² − 5x + 1x − 5 = x² − 4x − 5. ✓
x² − 4x − 5 = (x + 1)(x − 5)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x + 1)(x − 5) = 0
So (x − 5) = 0 or (x + 1) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 5 = 0
Add 5 to both sides: x = 5

x + 1 = 0
Subtract 1 from both sides: x = −1
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −1 or x = 5

Check. When x = −1: left side = (−1)² = 1 = 1; right side = 4 × (−1) + 5 = 1. Both sides equal. ✓
When x = 5: left side = 5² = 25 = 25; right side = 4 × 5 + 5 = 25. Both sides equal. ✓

Watch out. Do not divide both sides by x: you would lose the solution x = 0. And never set a bracket equal to a number other than 0.

Done

PSA number added to its square gives 42. Find the two possible numbers.

Method. To solve a quadratic equation by factorising: (1) get one side equal to 0, (2) factorise, (3) use the null factor law to make each bracket equal to 0, (4) solve the little equations.

Write the sentence as an equation. A number is x. Its square is x². Adding the number gives x² + x.
x² + x = 42
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 42 from both sides: x² + x − 42 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 1 and c = -42.
We need two numbers that multiply to -42 (the last number) and add to 1 (the middle number). List the pairs that multiply to -42:
1 × -42 = -42, add: 1 + -42 = -41 ✗
-1 × 42 = -42, add: -1 + 42 = 41 ✗
2 × -21 = -42, add: 2 + -21 = -19 ✗
-2 × 21 = -42, add: -2 + 21 = 19 ✗
3 × -14 = -42, add: 3 + -14 = -11 ✗
-3 × 14 = -42, add: -3 + 14 = 11 ✗
6 × -7 = -42, add: 6 + -7 = -1 ✗
-6 × 7 = -42, add: -6 + 7 = 1 ✓ this pair works
Put the pair into two brackets: (x − 6)(x + 7).
Check by expanding: x² + 7x − 6x − 42 = x² + x − 42. ✓
x² + x − 42 = (x − 6)(x + 7)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x − 6)(x + 7) = 0
So (x − 6) = 0 or (x + 7) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 6 = 0
Add 6 to both sides: x = 6

x + 7 = 0
Subtract 7 from both sides: x = −7
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −7 or x = 6

Check. When x = −7: left side = (−7)² + (−7) = 49 − 7 = 42; right side = 42 = 42. Both sides equal. ✓
When x = 6: left side = 6² + 6 = 36 + 6 = 42; right side = 42 = 42. Both sides equal. ✓

Watch out. Do not divide both sides by x: you would lose the solution x = 0. And never set a bracket equal to a number other than 0.

Done

PSA rectangle has width w cm and length (w + 5) cm. Its area is 14 cm². Find the width and the length.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
Let the width be w cm. Then the length is (w + 5) cm.
Form an equation from the information. Area of a rectangle = length × width.
w(w + 5) = 14
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand w(w + 5):
w × w = w²
w × 5 = 5w
so w(w + 5) = w² + 5w
The equation is now: w² + 5w = 14
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 14 from both sides: w² + 5w − 14 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form w² + bw + c, with b = 5 and c = -14.
We need two numbers that multiply to -14 (the last number) and add to 5 (the middle number). List the pairs that multiply to -14:
1 × -14 = -14, add: 1 + -14 = -13 ✗
-1 × 14 = -14, add: -1 + 14 = 13 ✗
2 × -7 = -14, add: 2 + -7 = -5 ✗
-2 × 7 = -14, add: -2 + 7 = 5 ✓ this pair works
Put the pair into two brackets: (w − 2)(w + 7).
Check by expanding: w² + 7w − 2w − 14 = w² + 5w − 14. ✓
w² + 5w − 14 = (w − 2)(w + 7)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(w − 2)(w + 7) = 0
So (w − 2) = 0 or (w + 7) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
w − 2 = 0
Add 2 to both sides: w = 2

w + 7 = 0
Subtract 7 from both sides: w = −7
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
w = −7 or w = 2
Decide which solution makes sense. w is a width, so it cannot be negative. Reject w = −7.
w = 2 cm
Find the length and answer the question. The length is w + 5.
Length = 2 + 5 = 7 cm
Width = 2 cm, length = 7 cm

Check. Check in the words: width 2 cm and length 7 cm differ by 5 cm and the area is 2 × 7 = 14 cm². ✓

Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).

Done

PSA rectangular garden has a perimeter of 34 m and an area of 60 m². Find the length and the width of the garden.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
Perimeter = 34 m, so length + width = 17 m. Let the length be l m. Then the width is (17 − l) m.
Form an equation from the information. Area = length × width = 60.
l(17 − l) = 60
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand l(−l + 17):
l × (−l) = −l²
l × 17 = 17l
so l(−l + 17) = −l² + 17l
The equation is now: −l² + 17l = 60
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Add l² to both sides: 17l = 60 + l²
Subtract 17l from both sides: 0 = 60 + l² − 17l
Collect like terms: 0 = l² − 17l + 60
Write it the other way round: l² − 17l + 60 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form l² + bl + c, with b = -17 and c = 60.
We need two numbers that multiply to 60 (the last number) and add to -17 (the middle number). List the pairs that multiply to 60:
1 × 60 = 60, add: 1 + 60 = 61 ✗
-1 × -60 = 60, add: -1 + -60 = -61 ✗
2 × 30 = 60, add: 2 + 30 = 32 ✗
-2 × -30 = 60, add: -2 + -30 = -32 ✗
3 × 20 = 60, add: 3 + 20 = 23 ✗
-3 × -20 = 60, add: -3 + -20 = -23 ✗
4 × 15 = 60, add: 4 + 15 = 19 ✗
-4 × -15 = 60, add: -4 + -15 = -19 ✗
5 × 12 = 60, add: 5 + 12 = 17 ✗
-5 × -12 = 60, add: -5 + -12 = -17 ✓ this pair works
Put the pair into two brackets: (l − 5)(l − 12).
Check by expanding: l² − 12l − 5l + 60 = l² − 17l + 60. ✓
l² − 17l + 60 = (l − 5)(l − 12)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(l − 5)(l − 12) = 0
So (l − 12) = 0 or (l − 5) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
l − 12 = 0
Add 12 to both sides: l = 12

l − 5 = 0
Add 5 to both sides: l = 5
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
l = 5 or l = 12
Decide which solutions make sense. Both answers are positive. They give the same rectangle, once as the length and once as the width.
l = 5 gives width 12 m; l = 12 gives width 5 m
Write the dimensions. The longer side is the length.
The garden is 12 m by 5 m.

Check. Perimeter = 2 × (12 + 5) = 34 m and area = 12 × 5 = 60 m². ✓

Watch out. When the equation is not in the form “x² + …”, rearrange it first: 17l − l² = 60 becomes l² − 17l + 60 = 0.

Done

PSSam is x years old. His sister is 3 years younger. The product of their ages is 130. Find their ages.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
Let Sam's age be x years. His sister is 3 years younger, so she is (x − 3) years old.
Form an equation from the information. The product of their ages is 130.
x(x − 3) = 130
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand x(x − 3):
x × x = x²
x × (−3) = −3x
so x(x − 3) = x² − 3x
The equation is now: x² − 3x = 130
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 130 from both sides: x² − 3x − 130 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = -3 and c = -130.
We need two numbers that multiply to -130 (the last number) and add to -3 (the middle number). List the pairs that multiply to -130:
1 × -130 = -130, add: 1 + -130 = -129 ✗
-1 × 130 = -130, add: -1 + 130 = 129 ✗
2 × -65 = -130, add: 2 + -65 = -63 ✗
-2 × 65 = -130, add: -2 + 65 = 63 ✗
5 × -26 = -130, add: 5 + -26 = -21 ✗
-5 × 26 = -130, add: -5 + 26 = 21 ✗
10 × -13 = -130, add: 10 + -13 = -3 ✓ this pair works
Put the pair into two brackets: (x + 10)(x − 13).
Check by expanding: x² − 13x + 10x − 130 = x² − 3x − 130. ✓
x² − 3x − 130 = (x + 10)(x − 13)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x + 10)(x − 13) = 0
So (x − 13) = 0 or (x + 10) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 13 = 0
Add 13 to both sides: x = 13

x + 10 = 0
Subtract 10 from both sides: x = −10
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −10 or x = 13
Decide which solution makes sense. An age cannot be negative, so reject x = −10.
x = 13
Find both ages. Sam is x and his sister is x − 3.
Sam: 13 years; sister: 13 − 3 = 10 years

Check. 13 × 10 = 130. ✓

Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).

Done

PSA triangle has base (x + 4) cm and height x cm. Its area is 30 cm². Find the base and the height.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
Let the height of the triangle be x cm and the base (x + 4) cm.
Form an equation from the information. Area = ½ × base × height = 30, so base × height = 60.
x(x + 4) = 60
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand x(x + 4):
x × x = x²
x × 4 = 4x
so x(x + 4) = x² + 4x
The equation is now: x² + 4x = 60
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 60 from both sides: x² + 4x − 60 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 4 and c = -60.
We need two numbers that multiply to -60 (the last number) and add to 4 (the middle number). List the pairs that multiply to -60:
1 × -60 = -60, add: 1 + -60 = -59 ✗
-1 × 60 = -60, add: -1 + 60 = 59 ✗
2 × -30 = -60, add: 2 + -30 = -28 ✗
-2 × 30 = -60, add: -2 + 30 = 28 ✗
3 × -20 = -60, add: 3 + -20 = -17 ✗
-3 × 20 = -60, add: -3 + 20 = 17 ✗
4 × -15 = -60, add: 4 + -15 = -11 ✗
-4 × 15 = -60, add: -4 + 15 = 11 ✗
5 × -12 = -60, add: 5 + -12 = -7 ✗
-5 × 12 = -60, add: -5 + 12 = 7 ✗
6 × -10 = -60, add: 6 + -10 = -4 ✗
-6 × 10 = -60, add: -6 + 10 = 4 ✓ this pair works
Put the pair into two brackets: (x − 6)(x + 10).
Check by expanding: x² + 10x − 6x − 60 = x² + 4x − 60. ✓
x² + 4x − 60 = (x − 6)(x + 10)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x − 6)(x + 10) = 0
So (x − 6) = 0 or (x + 10) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 6 = 0
Add 6 to both sides: x = 6

x + 10 = 0
Subtract 10 from both sides: x = −10
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −10 or x = 6
Decide which solution makes sense. Reject the negative height x = −10.
x = 6
Find the base. The base is x + 4.
Base = 6 + 4 = 10 cm. Height 6 cm, base 10 cm.

Check. ½ × 10 × 6 = 30. ✓

Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).

Done

PSA photograph measures 12 cm by 8 cm. It is put in a frame with a border of width x cm all round. The area of the whole frame, including the photograph, is 140 cm². Find x.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
The border has width x cm on every side. So the whole frame is (12 + 2x) cm long and (8 + 2x) cm wide (the border is added at both ends).
Form an equation from the information. The whole frame has area 140 cm², which is length × width.
(12 + 2x)(8 + 2x) = 140
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand (2x + 12)(2x + 8):
First: 2x × 2x = 4x²
Outer: 2x × 8 = 16x
Inner: 12 × 2x = 24x
Last: 12 × 8 = 96
so (2x + 12)(2x + 8) = 4x² + 40x + 96
The equation is now: 4x² + 40x + 96 = 140
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 140 from both sides: 4x² + 40x + 96 − 140 = 0
Collect like terms: 4x² + 40x − 44 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Look for a common factor. A common factor is something that divides into every term.
The terms are 4x², 40x, −44. The biggest number and the highest power of x that divide into all of them is 4.
Take it outside a bracket, and inside the bracket put each term divided by it:
4x² ÷ 4 = x²
40x ÷ 4 = 10x
−44 ÷ 4 = −11
So 4x² + 40x − 44 = 4(x² + 10x − 11).
Check by expanding: 4 × (x² + 10x − 11) = 4 × x², 4 × 10x, 4 × (−11) → 4x² + 40x − 44. ✓
Can the bracket be factorised again? Look at (x² + 10x − 11).
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = 10 and c = -11.
We need two numbers that multiply to -11 (the last number) and add to 10 (the middle number). List the pairs that multiply to -11:
1 × -11 = -11, add: 1 + -11 = -10 ✗
-1 × 11 = -11, add: -1 + 11 = 10 ✓ this pair works
Put the pair into two brackets: (x − 1)(x + 11).
Check by expanding: x² + 11x − 1x − 11 = x² + 10x − 11. ✓
4x² + 40x − 44 = 4(x − 1)(x + 11)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
4(x − 1)(x + 11) = 0
So (x − 1) = 0 or (x + 11) = 0.
The number 4 in front can never be 0, so it gives no solution. Ignore it.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 1 = 0
Add 1 to both sides: x = 1

x + 11 = 0
Subtract 11 from both sides: x = −11
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −11 or x = 1
Decide which solution makes sense. The border width cannot be negative, so reject x = −11.
x = 1
Answer in words. The border width is x.
The border is 1 cm wide.

Check. Whole frame: (12 + 2) by (8 + 2) = 14 by 10, and 14 × 10 = 140 cm². ✓

Watch out. Every term has a common factor of 4: dividing by 4 makes the factorising easier, but the answers are the same.

Done

PSSolve these problems about consecutive odd numbers. Form an equation first.

(a)The product of two consecutive odd positive whole numbers is 143. Find the numbers.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
Let the smaller number be n. The next consecutive odd number is n + 2 (odd numbers are 2 apart).
Form an equation from the information. The product of the two numbers means they are multiplied.
n(n + 2) = 143
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand n(n + 2):
n × n = n²
n × 2 = 2n
so n(n + 2) = n² + 2n
The equation is now: n² + 2n = 143
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 143 from both sides: n² + 2n − 143 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form n² + bn + c, with b = 2 and c = -143.
We need two numbers that multiply to -143 (the last number) and add to 2 (the middle number). List the pairs that multiply to -143:
1 × -143 = -143, add: 1 + -143 = -142 ✗
-1 × 143 = -143, add: -1 + 143 = 142 ✗
11 × -13 = -143, add: 11 + -13 = -2 ✗
-11 × 13 = -143, add: -11 + 13 = 2 ✓ this pair works
Put the pair into two brackets: (n − 11)(n + 13).
Check by expanding: n² + 13n − 11n − 143 = n² + 2n − 143. ✓
n² + 2n − 143 = (n − 11)(n + 13)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(n − 11)(n + 13) = 0
So (n − 11) = 0 or (n + 13) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
n − 11 = 0
Add 11 to both sides: n = 11

n + 13 = 0
Subtract 13 from both sides: n = −13
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
n = −13 or n = 11
Decide which solution makes sense. The numbers are positive, so reject n = −13.
n = 11
Find both numbers. The numbers are n and n + 2.
11 and 11 + 2 = 13
The numbers are 11 and 13.

Check. Check: 11 × 13 = 143. ✓

Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).

(b)The product of two consecutive odd positive whole numbers is 195. Find the numbers.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
Let the smaller number be n. The next consecutive odd number is n + 2 (odd numbers are 2 apart).
Form an equation from the information. The product of the two numbers means they are multiplied.
n(n + 2) = 195
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand n(n + 2):
n × n = n²
n × 2 = 2n
so n(n + 2) = n² + 2n
The equation is now: n² + 2n = 195
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 195 from both sides: n² + 2n − 195 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form n² + bn + c, with b = 2 and c = -195.
We need two numbers that multiply to -195 (the last number) and add to 2 (the middle number). List the pairs that multiply to -195:
1 × -195 = -195, add: 1 + -195 = -194 ✗
-1 × 195 = -195, add: -1 + 195 = 194 ✗
3 × -65 = -195, add: 3 + -65 = -62 ✗
-3 × 65 = -195, add: -3 + 65 = 62 ✗
5 × -39 = -195, add: 5 + -39 = -34 ✗
-5 × 39 = -195, add: -5 + 39 = 34 ✗
13 × -15 = -195, add: 13 + -15 = -2 ✗
-13 × 15 = -195, add: -13 + 15 = 2 ✓ this pair works
Put the pair into two brackets: (n − 13)(n + 15).
Check by expanding: n² + 15n − 13n − 195 = n² + 2n − 195. ✓
n² + 2n − 195 = (n − 13)(n + 15)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(n − 13)(n + 15) = 0
So (n − 13) = 0 or (n + 15) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
n − 13 = 0
Add 13 to both sides: n = 13

n + 15 = 0
Subtract 15 from both sides: n = −15
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
n = −15 or n = 13
Decide which solution makes sense. The numbers are positive, so reject n = −15.
n = 13
Find both numbers. The numbers are n and n + 2.
13 and 13 + 2 = 15
The numbers are 13 and 15.

Check. Check: 13 × 15 = 195. ✓

Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).

(c)The product of two consecutive odd positive whole numbers is 99. Find the numbers.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
Let the smaller number be n. The next consecutive odd number is n + 2 (odd numbers are 2 apart).
Form an equation from the information. The product of the two numbers means they are multiplied.
n(n + 2) = 99
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand n(n + 2):
n × n = n²
n × 2 = 2n
so n(n + 2) = n² + 2n
The equation is now: n² + 2n = 99
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 99 from both sides: n² + 2n − 99 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form n² + bn + c, with b = 2 and c = -99.
We need two numbers that multiply to -99 (the last number) and add to 2 (the middle number). List the pairs that multiply to -99:
1 × -99 = -99, add: 1 + -99 = -98 ✗
-1 × 99 = -99, add: -1 + 99 = 98 ✗
3 × -33 = -99, add: 3 + -33 = -30 ✗
-3 × 33 = -99, add: -3 + 33 = 30 ✗
9 × -11 = -99, add: 9 + -11 = -2 ✗
-9 × 11 = -99, add: -9 + 11 = 2 ✓ this pair works
Put the pair into two brackets: (n − 9)(n + 11).
Check by expanding: n² + 11n − 9n − 99 = n² + 2n − 99. ✓
n² + 2n − 99 = (n − 9)(n + 11)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(n − 9)(n + 11) = 0
So (n − 9) = 0 or (n + 11) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
n − 9 = 0
Add 9 to both sides: n = 9

n + 11 = 0
Subtract 11 from both sides: n = −11
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
n = −11 or n = 9
Decide which solution makes sense. The numbers are positive, so reject n = −11.
n = 9
Find both numbers. The numbers are n and n + 2.
9 and 9 + 2 = 11
The numbers are 9 and 11.

Check. Check: 9 × 11 = 99. ✓

Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).

Done

PSThe sum of the squares of two consecutive whole numbers is 85. Find the two numbers. (There are two possible pairs.)

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
Let the smaller number be n. The next consecutive whole number is n + 1.
Form an equation from the information. The sum of the squares of the two numbers is 85.
n² + (n + 1)² = 85
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand (n + 1)(n + 1):
First: n × n = n²
Outer: n × 1 = n
Inner: 1 × n = n
Last: 1 × 1 = 1
so (n + 1)(n + 1) = n² + 2n + 1
The equation is now: 2n² + 2n + 1 = 85
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 85 from both sides: 2n² + 2n + 1 − 85 = 0
Collect like terms: 2n² + 2n − 84 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Look for a common factor. A common factor is something that divides into every term.
The terms are 2n², 2n, −84. The biggest number and the highest power of n that divide into all of them is 2.
Take it outside a bracket, and inside the bracket put each term divided by it:
2n² ÷ 2 = n²
2n ÷ 2 = n
−84 ÷ 2 = −42
So 2n² + 2n − 84 = 2(n² + n − 42).
Check by expanding: 2 × (n² + n − 42) = 2 × n², 2 × n, 2 × (−42) → 2n² + 2n − 84. ✓
Can the bracket be factorised again? Look at (n² + n − 42).
Spot the pattern. This is a quadratic of the form n² + bn + c, with b = 1 and c = -42.
We need two numbers that multiply to -42 (the last number) and add to 1 (the middle number). List the pairs that multiply to -42:
1 × -42 = -42, add: 1 + -42 = -41 ✗
-1 × 42 = -42, add: -1 + 42 = 41 ✗
2 × -21 = -42, add: 2 + -21 = -19 ✗
-2 × 21 = -42, add: -2 + 21 = 19 ✗
3 × -14 = -42, add: 3 + -14 = -11 ✗
-3 × 14 = -42, add: -3 + 14 = 11 ✗
6 × -7 = -42, add: 6 + -7 = -1 ✗
-6 × 7 = -42, add: -6 + 7 = 1 ✓ this pair works
Put the pair into two brackets: (n − 6)(n + 7).
Check by expanding: n² + 7n − 6n − 42 = n² + n − 42. ✓
2n² + 2n − 84 = 2(n − 6)(n + 7)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
2(n − 6)(n + 7) = 0
So (n − 6) = 0 or (n + 7) = 0.
The number 2 in front can never be 0, so it gives no solution. Ignore it.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
n − 6 = 0
Add 6 to both sides: n = 6

n + 7 = 0
Subtract 7 from both sides: n = −7
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
n = −7 or n = 6
Decide which solutions make sense. The question says only “whole numbers”, not “positive”. So both solutions are allowed.
n = −7 gives the pair −7 and −6
n = 6 gives the pair 6 and 7
Answer in words. There are two possible pairs.
The numbers are 6 and 7, or −7 and −6.

Check. 6² + 7² = 36 + 49 = 85 and (−7)² + (−6)² = 49 + 36 = 85. ✓

Watch out. Only reject a negative answer if the question says the numbers are positive.

Done

PSA rectangle has length (2x + 1) cm and width (x − 1) cm. Its area is 27 cm². Find x and write down the dimensions of the rectangle.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
Let the width be (x − 1) cm and the length (2x + 1) cm, as in the diagram.
Form an equation from the information. Area = length × width = 27.
(2x + 1)(x − 1) = 27
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand (2x + 1)(x − 1):
First: 2x × x = 2x²
Outer: 2x × (−1) = −2x
Inner: 1 × x = x
Last: 1 × (−1) = −1
so (2x + 1)(x − 1) = 2x² − x − 1
The equation is now: 2x² − x − 1 = 27
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 27 from both sides: 2x² − x − 1 − 27 = 0
Collect like terms: 2x² − x − 28 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. A quadratic where the number in front of x² is 2, not 1. We use the split the middle term method.
Multiply the first and last numbers: 2 × (-28) = -56. We need two numbers that multiply to -56 and add to -1 (the middle number).
1 × -56 = -56, add: -55 ✗
-1 × 56 = -56, add: 55 ✗
2 × -28 = -56, add: -26 ✗
-2 × 28 = -56, add: 26 ✗
4 × -14 = -56, add: -10 ✗
-4 × 14 = -56, add: 10 ✗
7 × -8 = -56, add: -1 ✓ this pair works
Split the middle term (-1x) into 7x and -8x: 2x² + 7x − 8x − 28.
Factorise each pair separately. First pair: 2x² + 7x = x(2x + 7). Second pair: −8x − 28 = −4(2x + 7).
The two brackets (2x + 7) are the same. That is how we know it has worked. Write: x(2x + 7) − 4(2x + 7).
Take the common bracket out: (2x + 7) multiplied by (x − 4).
Result: (x − 4)(2x + 7).
Check by expanding: (x − 4)(2x + 7) = 2x² − x − 28. ✓
2x² − x − 28 = (x − 4)(2x + 7)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x − 4)(2x + 7) = 0
So (x − 4) = 0 or (2x + 7) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 4 = 0
Add 4 to both sides: x = 4

2x + 7 = 0
Subtract 7 from both sides: 2x = −7
Divide both sides by 2: x = −7 ÷ 2 = −72
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −72 or x = 4
Decide which solution makes sense. Reject x = −7/2 because it would make the width x − 1 negative.
x = 4
Find the length and the width. Put x = 4 into each expression.
Width = 4 − 1 = 3 cm
Length = 2 × 4 + 1 = 9 cm
The rectangle is 9 cm by 3 cm.

Check. 9 × 3 = 27. ✓

Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).

Done

PSThe three sides of a right-angled triangle are x cm, (x + 7) cm and (x + 8) cm. Find the length of each side.

Method. Turn the words into a quadratic equation, solve it by factorising, and then decide which solutions make sense in the context.

Choose a letter for the unknown. Algebra needs a letter to stand for the thing we do not know. Say clearly what it stands for, including the units.
The shortest side is x cm, the middle side is (x + 7) cm and the hypotenuse is (x + 8) cm.
Form an equation from the information. Pythagoras: (shorter side)² + (other side)² = (hypotenuse)².
x² + (x + 7)² = (x + 8)²
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand (x + 7)(x + 7):
First: x × x = x²
Outer: x × 7 = 7x
Inner: 7 × x = 7x
Last: 7 × 7 = 49
so (x + 7)(x + 7) = x² + 14x + 49

Expand (x + 8)(x + 8):
First: x × x = x²
Outer: x × 8 = 8x
Inner: 8 × x = 8x
Last: 8 × 8 = 64
so (x + 8)(x + 8) = x² + 16x + 64
The equation is now: 2x² + 14x + 49 = x² + 16x + 64
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract x² from both sides: 2x² + 14x + 49 − x² = 16x + 64
Subtract 16x from both sides: 2x² + 14x + 49 − x² − 16x = 64
Subtract 64 from both sides: 2x² + 14x + 49 − x² − 16x − 64 = 0
Collect like terms: x² − 2x − 15 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Spot the pattern. This is a quadratic of the form x² + bx + c, with b = -2 and c = -15.
We need two numbers that multiply to -15 (the last number) and add to -2 (the middle number). List the pairs that multiply to -15:
1 × -15 = -15, add: 1 + -15 = -14 ✗
-1 × 15 = -15, add: -1 + 15 = 14 ✗
3 × -5 = -15, add: 3 + -5 = -2 ✓ this pair works
Put the pair into two brackets: (x + 3)(x − 5).
Check by expanding: x² − 5x + 3x − 15 = x² − 2x − 15. ✓
x² − 2x − 15 = (x + 3)(x − 5)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
(x + 3)(x − 5) = 0
So (x − 5) = 0 or (x + 3) = 0.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
x − 5 = 0
Add 5 to both sides: x = 5

x + 3 = 0
Subtract 3 from both sides: x = −3
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
x = −3 or x = 5
Decide which solution makes sense. A length cannot be negative: reject x = −3.
x = 5
Find the three sides. Put x = 5 into each expression.
Sides: 5 cm, 5 + 7 = 12 cm, 5 + 8 = 13 cm

Check. 5² + 12² = 25 + 144 = 169 = 13². ✓

Watch out. In a problem, a mathematically correct solution might not make sense: always ask “is this answer possible?” (a length cannot be negative).

Done

RThe sum of the squares of three consecutive whole numbers is 365. Find the three numbers. (There are two possible sets.)

Method. To solve a quadratic equation by factorising: (1) get one side equal to 0, (2) factorise, (3) use the null factor law to make each bracket equal to 0, (4) solve the little equations.

Choose a letter. Let the middle number be n. Then the three consecutive numbers are n − 1, n and n + 1.
Numbers: n − 1, n, n + 1
Form the equation. The sum of the squares of the three numbers is 365. Expanding the squares: (n − 1)² = n² − 2n + 1 and (n + 1)² = n² + 2n + 1. The 2n terms cancel, so the left side is 3n² + 2.
(n − 1)² + n² + (n + 1)² = 365
3n² + 2 = 365
Expand any brackets first. Multiply every term in one bracket by every term in the other. We cannot see how to factorise until the brackets are gone.
Expand (n − 1)(n − 1):
First: n × n = n²
Outer: n × (−1) = −n
Inner: (−1) × n = −n
Last: (−1) × (−1) = 1
so (n − 1)(n − 1) = n² − 2n + 1

Expand (n + 1)(n + 1):
First: n × n = n²
Outer: n × 1 = n
Inner: 1 × n = n
Last: 1 × 1 = 1
so (n + 1)(n + 1) = n² + 2n + 1
The equation is now: 3n² + 2 = 365
Rearrange so that one side equals 0. A quadratic must be rearranged so that one side is exactly 0. The null factor law (see below) only works on a product that equals 0. Move every term to the side where the squared term is positive, using inverse operations on both sides.
Subtract 365 from both sides: 3n² + 2 − 365 = 0
Collect like terms: 3n² − 363 = 0
Factorise the quadratic. Factorising writes it as brackets multiplied together. Then the null factor law can be used.
Look for a common factor. A common factor is something that divides into every term.
The terms are 3n², −363. The biggest number and the highest power of n that divide into all of them is 3.
Take it outside a bracket, and inside the bracket put each term divided by it:
3n² ÷ 3 = n²
−363 ÷ 3 = −121
So 3n² − 363 = 3(n² − 121).
Check by expanding: 3 × (n² − 121) = 3 × n², 3 × (−121) → 3n² − 363. ✓
Can the bracket be factorised again? Look at (n² − 121).
Spot the pattern. There are two terms, both are perfect squares, and there is a minus between them. This is a difference of two squares.
The rule is: a² − b² = (a − b)(a + b).
Find a and b. n² = (n)², so a = n. And 121 = 11², so b = 11.
Put them into the rule: (n − 11)(n + 11).
Check by expanding: (n − 11)(n + 11) = n² + 11n − 11n − 121 = n² − 121. ✓ (the middle terms cancel)
3n² − 363 = 3(n − 11)(n + 11)
Use the null factor law. The null factor law. If two numbers multiply to make 0, then at least one of them must be 0. Think about it: 7 × 0 = 0 and 0 × 12 = 0, but 7 × 12 is not 0. The same is true when the numbers are brackets: if (x + 2)(x − 5) = 0, then either the first bracket is 0 or the second bracket is 0.
3(n − 11)(n + 11) = 0
So (n − 11) = 0 or (n + 11) = 0.
The number 3 in front can never be 0, so it gives no solution. Ignore it.
Solve each bracket separately. Each bracket gives a very small equation of the kind you solved in Chapter 9. Use inverse operations on both sides.
n − 11 = 0
Add 11 to both sides: n = 11

n + 11 = 0
Subtract 11 from both sides: n = −11
Write the solutions. Quadratics usually have two solutions (a curve crosses the x-axis twice).
n = −11 or n = 11
Decide which solutions make sense. The question says only “whole numbers”, so both solutions are allowed. The middle number is n.
n = 11 gives 10, 11, 12
n = −11 gives −12, −11, −10

Check. 10² + 11² + 12² = 100 + 121 + 144 = 365. ✓ And (−12)² + (−11)² + (−10)² = 144 + 121 + 100 = 365. ✓

Watch out. Do not divide both sides by x: you would lose the solution x = 0. And never set a bracket equal to a number other than 0.

Done

RExplain why two consecutive whole numbers cannot have a product of 50.

Done