Websink Prep UKKS3 MathsYear 9 chaptersMaths textbooksReceptionYear 1Year 2Year 4Year 5Year 6Year 3Year 7Year 8Year 9GCSEBrowse allPast papersFind a tutorBecome a tutor
KS3 Maths · Year 9Chapter 4 · Quadratic expressions and equations
Year 9 · Chapter 4

Quadratic expressions and equations

By the end of this chapter you will be able to

  • Expand two brackets, including brackets with a coefficient and squared brackets, and simplify
  • Factorise ax2 + bx + c using the ac method, and recognise a difference of two squares
  • Solve quadratic equations by factorising, including x2 = kx and equations with fractional solutions
  • Extension: complete the square and use the quadratic formula to give answers in surd form
  • Form a quadratic equation from a rectangle or from consecutive numbers, solve it and reject an unsuitable answer

You should already know

  • Expanding and simplifying a single bracket and collecting like terms (Year 8, Chapter 5)
  • Factorising x2 + bx + c with integers (Year 8, Chapter 5)
  • Solving linear equations and checking a solution (Year 8, Chapter 6)
  • Simplifying surds such as √50 = 5√2 (Year 9, Chapter 1)

Where this leads

  • Graphs of quadratic functions and their roots (Year 9, Chapter 6)
  • Simultaneous equations with a quadratic (Year 9, Chapter 5 extension)
  • Quadratic equations and graphs in GCSE algebra

National Curriculum for England, Key Stage 3 mathematics, Year 9. Content: Expand double brackets, including squares and products such as (2x − 1)(x + 4), and simplify, Factorise quadratics of the form ax2 + bx + c, and the difference of two squares such as x2 − 9, Solve quadratic equations by factorising, using the zero product property, Extension: solve quadratic equations by completing the square or by the quadratic formula, Form and solve a quadratic equation from a context, such as the sides of a rectangle with a given area.

Exercise P  Do you remember?

Non-calculator
1
FExpand 2(x + 3).
2
FSolve 2x + 3 = 11.
3
FWork out (−3)2.
4
FA rectangle is 4 cm by 7 cm. Work out its area.
5
FFactorise x2 + 5x + 6.
4.1

Expanding double brackets

Year 9
Expand double brackets, including squares, and simplify
Video lesson: Expanding double bracketsComing soon
A square garden has sides of (x + 3) metres. To find its area, Priya multiplies (x + 3) by (x + 3). Multiplying two brackets together gives a quadratic expression with an x2 term.

To expand two brackets, multiply every term in the first bracket by every term in the second. Then collect the like terms. For (px + q)(rx + s) the four products are px × rx, px × s, q × rx and q × s.

A squared bracket is a bracket multiplied by itself: (x + 3)2 = (x + 3)(x + 3).

Key ideas

  • Multiply every term in one bracket by every term in the other.
  • x × x = x2, and a negative times a negative is a positive.
  • Collect the like terms: the two x terms add together.
  • (x + a)2 = x2 + 2ax + a2. Do not write x2 + a2.
Worked example 4.1.1
Expand and simplify (x + 3)2.

Method. Multiply every term in one bracket by every term in the other, then collect like terms.

  1. Multiply the first terms. The x in each bracket multiplies to give x2.
    (x) × (x) = x2
  2. Multiply the outside terms. The first bracket times the constant of the second.
    (x) × (3) = 3x
  3. Multiply the inside terms. The constant of the first bracket times the x of the second.
    (3) × (x) = 3x
  4. Multiply the last terms. The two constants multiply together.
    (3) × (3) = 9
  5. Collect the like terms. Add the two x terms together.
    3x + 3x = 6x
  6. Write the answer.
    x2 + 6x + 9

Check. Check with x = 1: the brackets give (4)(4) = 16, and the expanded form also gives 16.

Watch out. Check each product carefully. The most common error is forgetting the sign of the constant.

Worked example 4.1.2
Expand and simplify (2x + 1)(x + 4).

Method. Multiply every term in one bracket by every term in the other, then collect like terms.

  1. Multiply the first terms. The x in each bracket multiplies to give x2.
    (2x) × (x) = 2x2
  2. Multiply the outside terms. The first bracket times the constant of the second.
    (2x) × (4) = 8x
  3. Multiply the inside terms. The constant of the first bracket times the x of the second.
    (1) × (x) = x
  4. Multiply the last terms. The two constants multiply together.
    (1) × (4) = 4
  5. Collect the like terms. Add the two x terms together.
    8x + x = 9x
  6. Write the answer.
    2x2 + 9x + 4

Check. Check with x = 1: the brackets give (3)(5) = 15, and the expanded form also gives 15.

Watch out. Check each product carefully. The most common error is forgetting the sign of the constant.

Worked example 4.1.3
Expand and simplify (x + 4)(x − 4).

Method. Multiply every term in one bracket by every term in the other, then collect like terms.

  1. Multiply the first terms. The x in each bracket multiplies to give x2.
    (x) × (x) = x2
  2. Multiply the outside terms. The first bracket times the constant of the second.
    (x) × (−4) = −4x
  3. Multiply the inside terms. The constant of the first bracket times the x of the second.
    (4) × (x) = 4x
  4. Multiply the last terms. The two constants multiply together.
    (4) × (−4) = −16
  5. Write the answer.
    x2 − 16

Check. Check with x = 1: the brackets give (5)(−3) = −15, and the expanded form also gives −15.

Watch out. Check each product carefully. The most common error is forgetting the sign of the constant.

Watch out

Watch out: (x + 3)2 is not x2 + 9. The middle term 6x is missing from that answer.

Exam tip

Exam tip. Write out the four products on separate lines before you collect the terms.

Exercise 4A  Expanding double brackets

Non-calculator
1
FExpand and simplify (x + 3)2.
2
FExpand and simplify (x − 6)2.
3
FExpand and simplify (2x − 5)2.
4
FExpand and simplify (3x + 4)2.
5
FExpand and simplify (2x + 1)(x + 4).
6
FExpand and simplify (3x − 1)(x + 2).
7
FExpand and simplify (2x − 5)(3x + 1).
8
FExpand and simplify (x + 4)(x − 4).
9
PSExpand and simplify (5x + 2)(5x − 2).
10
PSExpand and simplify (4x − 1)(2x + 3).
11
PSExpand and simplify (x − 7)(2x + 5).
12
PSExpand and simplify (3x + 2)(−x + 5).
13
PSExpand and simplify (2x + 3)(4x − 1).
14
RExplain why (x + 5)2 is not the same as x2 + 25. Expand (x + 5)2 to show the correct answer.
4.2

Factorising quadratics

Year 9
Factorise ax2 + bx + c, and the difference of two squares
Video lesson: Factorising quadraticsComing soon
A rectangle has an area of 2x2 + 7x + 3 square centimetres. Its sides are two brackets, and finding them means factorising the quadratic expression.

To factorise ax2 + bx + c, use the ac method. Multiply a and c. Find two numbers that multiply to give ac and add to give b. Split the middle term with those two numbers, then factorise in two pairs. Both pairs share a bracket, which is the answer.

A difference of two squares such as x2 − 9 has no middle term. It factorises into (x − 3)(x + 3), because a2 − b2 = (a − b)(a + b).

Key ideas

  • ac method: multiply a and c, find two numbers that multiply to ac and add to b.
  • Split the middle term, then factorise in two pairs.
  • a2 − b2 = (a − b)(a + b).
  • Check your answer by expanding the brackets.
Worked example 4.2.1
Factorise 2x2 + 7x + 3.

Method. Use the ac method: find two numbers, split the middle term, then factorise in pairs.

  1. Multiply the number in front of x2 by the constant. The ac method starts by finding a × c.
    2 × 3 = 6
  2. Find two numbers that multiply to give a × c and add to give b. Try pairs of factors of the product until the sum is b.
    6 × 1 = 6, and 6 + 1 = 7
  3. Split the middle term. Write 7x as 6x + x.
    2x2 + 7x + 3 = 2x2 + 6x + 1x + 3
  4. Factorise the first pair. Take out the common factor of the first two terms.
    2x2 + 6x = 2x(x + 3)
  5. Factorise the second pair. Take out the common factor of the last two terms. The bracket must match the first one.
    x + 3 = 1(x + 3)
  6. Write the answer. Both pairs now share the same bracket, so take it out.
    (2x + 1)(x + 3)

Check. Multiply out (2x + 1)(x + 3): the x2 term is 2, the x term is 7 and the constant is 3, so it is correct.

Watch out. Check the signs. A negative constant means the brackets have opposite signs.

Worked example 4.2.2
Factorise 2x2 + 5x − 12.

Method. Use the ac method: find two numbers, split the middle term, then factorise in pairs.

  1. Multiply the number in front of x2 by the constant. The ac method starts by finding a × c.
    2 × -12 = -24
  2. Find two numbers that multiply to give a × c and add to give b. Try pairs of factors of the product until the sum is b.
    8 × -3 = -24, and 8 + -3 = 5
  3. Split the middle term. Write 5x as 8x + −3x.
    2x2 + 5x − 12 = 2x2 + 8x − 3x − 12
  4. Factorise the first pair. Take out the common factor of the first two terms.
    2x2 + 8x = 2x(x + 4)
  5. Factorise the second pair. Take out the common factor of the last two terms. The bracket must match the first one.
    −3x − 12 = −3(x + 4)
  6. Write the answer. Both pairs now share the same bracket, so take it out.
    (2x − 3)(x + 4)

Check. Multiply out (2x − 3)(x + 4): the x2 term is 2, the x term is 5 and the constant is -12, so it is correct.

Watch out. Check the signs. A negative constant means the brackets have opposite signs.

Worked example 4.2.3
Factorise x2 − 9.

Method. The difference of two squares factorises into one bracket with a minus sign and one with a plus sign.

  1. Write both terms as squares. 1x2 is (1x)2, and 9 is 32.
    x2 − 9 = (1x)2 − 32
  2. Use the difference of two squares. a2 − b2 = (a − b)(a + b).
    (1x)2 − 32 = (1x − 3)(1x + 3)
  3. Write the answer.
    (x − 3)(x + 3)

Check. Multiply out: (1x − 3)(1x + 3) = 1x2 − 9.

Watch out. A sum of two squares does not factorise like this. Only a difference does.

Watch out

Watch out: a sum of two squares, such as x2 + 9, does not factorise into real brackets. Only a difference of two squares does.

Exam tip

Exam tip. Check the signs of the constants at the end. They tell you whether the brackets have the same or opposite signs.

Exercise 4B  Factorising quadratics

Non-calculator
1
FFactorise 2x2 + 7x + 3.
2
FFactorise 3x2 + 5x + 2.
3
FFactorise 2x2 + 5x − 12.
4
FFactorise 5x2 − 13x − 6.
5
FFactorise 4x2 − 7x − 2.
6
FFactorise 6x2 + 11x − 10.
7
PSFactorise 6x2 − 11x + 4.
8
PSFactorise 3x2 + 14x − 5.
9
PSFactorise 6x2 + 11x + 4.
10
FFactorise x2 − 9.
11
FFactorise 4x2 − 25.
12
FFactorise 9x2 − 16.
13
FFactorise x2 − 49.
14
PSFactorise 25x2 − 1.
4.3

Solving quadratic equations by factorising

Year 9
Solve quadratic equations by factorising, using the zero product property
Video lesson: Solving quadratic equations by factorisingComing soon
A ball is thrown and its height at time t is given by a quadratic. The times when the ball is at ground level are the solutions of a quadratic equation. To find them, factorise and use the fact that if two things multiply to zero, one of them is zero.

The zero product property says: if A × B = 0, then A = 0 or B = 0. So to solve a quadratic in factor form, set each bracket equal to zero and solve each linear equation.

If there is no constant term, such as x2 − 7x = 0, take out the x first: x(x − 7) = 0. One solution is x = 0.

Key ideas

  • Write the quadratic in factor form, with zero on one side.
  • Set each bracket equal to zero and solve each one.
  • Do not divide both sides by x. That loses the solution x = 0.
  • A quadratic usually has two solutions. Give both.
Worked example 4.3.1
Solve (x + 2)(x − 5) = 0.

Method. Solve by factorising: set each bracket equal to zero, then solve.

  1. Write the equation in factor form. The product of the brackets is zero.
    (x + 2)(x − 5) = 0
  2. Set the first bracket to zero. If one bracket is zero, the product is zero.
    x + 2 = 0, so x = −2
  3. Set the second bracket to zero. Solve the second bracket in the same way.
    x − 5 = 0, so x = 5
  4. Write the answer. Give both solutions.
    x = −2 or x = 5

Check. Substitute x = −2 and x = 5 into the original equation. Each gives 0.

Watch out. Solve each bracket separately. Do not divide both sides by x: that loses the solution x = 0.

Worked example 4.3.2
Solve (2x + 3)(x − 4) = 0.

Method. Solve by factorising: set each bracket equal to zero, then solve.

  1. Write the equation in factor form. The product of the brackets is zero.
    (2x + 3)(x − 4) = 0
  2. Set the first bracket to zero. If one bracket is zero, the product is zero.
    2x + 3 = 0, so x = −3/2
  3. Set the second bracket to zero. Solve the second bracket in the same way.
    x − 4 = 0, so x = 4
  4. Write the answer. Give both solutions.
    x = −3/2 or x = 4

Check. Substitute x = −3/2 and x = 4 into the original equation. Each gives 0.

Watch out. Solve each bracket separately. Do not divide both sides by x: that loses the solution x = 0.

Worked example 4.3.3
Solve x2 − 7x = 0.

Method. Factorise out the x, then solve each factor.

  1. Take out the common factor x. Both terms have an x, so write the equation as x times a bracket.
    x2 − 7x = x(x − 7)
  2. Set each factor to zero. Either x = 0 or the bracket is zero.
    x = 0 or x − 7 = 0
  3. Solve the bracket.
    x = 7
  4. Write the answer.
    x = 0 or x = 7

Check. Both 0 and 7 give 0 when substituted.

Watch out. Do not divide by x. The answer x = 0 is one of the solutions.

Watch out

Watch out: do not write x2 = x and then divide by x. You lose the solution x = 0.

Exam tip

Exam tip. Check each solution by substituting it back into the original equation.

Exercise 4C  Solving quadratic equations by factorising

Non-calculator
1
FSolve (x + 2)(x − 5) = 0.
2
FSolve (x + 3)(x + 1) = 0.
3
FSolve (x − 4)(x − 6) = 0.
4
FSolve (x − 6)(x + 6) = 0.
5
FSolve (2x + 3)(x − 4) = 0.
6
FSolve (3x − 2)(x − 5) = 0.
7
PSSolve (4x − 1)(2x + 3) = 0.
8
PSSolve (2x − 5)(x − 3) = 0.
9
FSolve x2 − 7x = 0.
10
FSolve x2 + 4x = 0.
11
FSolve x2 − 9x = 0.
12
PSSolve x2 + 5x + 6 = 0 by factorising first.
13
PSSolve x2 − 9 = 0.
14
RAli says that x2 − 4x = 0 has only one solution, x = 4. Is Ali right? Explain your answer and give all the solutions.
4.4

Extension: completing the square and the quadratic formula

Year 9
Extension: solve quadratics by completing the square or the quadratic formula
Video lesson: Extension: completing the square and the quadratic formulaComing soon
Some quadratics do not factorise with whole numbers. A garden design needs an exact length, so the answer is given as a surd, using a method that works for every quadratic.

Completing the square writes x2 + bx + c as (x + b/2)2 − (b/2)2 + c, which makes the equation easy to solve by taking square roots.

The quadratic formula gives the solutions of ax2 + bx + c = 0 for any a, b and c: x = (−b ± √(b2 − 4ac)) / (2a). The part under the square root is the discriminant.

Key ideas

  • Extension: this is more advanced than most KS3 work.
  • Completing the square: move the constant across, then take the square root of both sides, remembering ±.
  • Quadratic formula: substitute a, b and c, then simplify the surd.
  • Give answers in exact surd form unless the question asks for a decimal.
Worked example 4.4.1
Solve x2 + 6x + 2 = 0 by completing the square. Give your answer in surd form.

Method. Complete the square: write the x terms as a square, then take the square root of both sides.

  1. Write the x terms as a square. Half of the number in front of x goes inside the bracket.
    x2 + 6x = (x + 3)2 − 9
  2. Rewrite the whole equation. The square has to be corrected by taking away the square of the half.
    (x + 3)2 − 9 + 2 = 0
  3. Move the number to the other side. Add the number to both sides.
    (x + 3)2 = 7
  4. Take the square root of both sides. Remember both the positive and negative square roots.
    x + 3 = ±√7
  5. Solve for x. Move the number back to the other side.
    x = −3 ± √7

Check. Numerically, the two solutions are about -0.354 and -5.646.

Watch out. Do not forget the ±: a square root gives two answers.

Worked example 4.4.2
Solve 2x2 + 3x − 4 = 0 using the quadratic formula. Give your answer in surd form.

Method. Use the quadratic formula with the values of a, b and c.

  1. Write down a, b and c. The equation is in the form ax2 + bx + c = 0.
    a = 2, b = 3, c = -4
  2. Work out the discriminant. The discriminant is b2 − 4ac. It sits under the square root.
    32 − 4 × 2 × -4 = 9 − -32 = 41
  3. Substitute into the formula. x = (−b ± √(b2 − 4ac)) / (2a).
    x = (−(3) ± √41) / 4
  4. Simplify the surd. Take out the largest square factor of the discriminant.
    √41 = √41
  5. Write the two solutions. Use the + and − versions of the surd, then simplify each.
    x = (−3 + √41)/4 or x = (−3 − √41)/4

Check. Numerically the two solutions are about 0.851 and -2.351.

Watch out. Put brackets around the negative b: it is −(−3) = +3, not −3.

Worked example 4.4.3
Solve x2 − 4x − 1 = 0 by completing the square. Give your answer in surd form.

Method. Complete the square: write the x terms as a square, then take the square root of both sides.

  1. Write the x terms as a square. Half of the number in front of x goes inside the bracket.
    x2 − 4x = (x − 2)2 − 4
  2. Rewrite the whole equation. The square has to be corrected by taking away the square of the half.
    (x − 2)2 − 4 − 1 = 0
  3. Move the number to the other side. Add the number to both sides.
    (x − 2)2 = 5
  4. Take the square root of both sides. Remember both the positive and negative square roots.
    x − 2 = ±√5
  5. Solve for x. Move the number back to the other side.
    x = 2 ± √5

Check. Numerically, the two solutions are about 4.236 and -0.236.

Watch out. Do not forget the ±: a square root gives two answers.

Watch out

Watch out: in the quadratic formula the minus sign applies to b, so if b is negative, −b is positive. Use brackets.

Exam tip

Exam tip. Write down a, b and c before you substitute. Then check your answer by substituting it into the equation.

Exercise 4D  Completing the square and the quadratic formula (extension)

Non-calculator
1
FSolve x2 + 6x + 2 = 0 by completing the square. Give your answer in surd form.
2
FSolve x2 + 4x − 1 = 0 by completing the square. Give your answer in surd form.
3
PSSolve x2 − 6x + 4 = 0 by completing the square. Give your answer in surd form.
4
PSSolve x2 + 8x + 3 = 0 by completing the square. Give your answer in surd form.
5
PSSolve x2 − 4x − 1 = 0 by completing the square. Give your answer in surd form.
6
PSSolve x2 + 2x − 4 = 0 by completing the square. Give your answer in surd form.
7
FSolve 2x2 + 3x − 4 = 0 using the quadratic formula. Give your answer in surd form.
8
FSolve x2 − 5x + 2 = 0 using the quadratic formula. Give your answer in surd form.
9
PSSolve 3x2 + 2x − 2 = 0 using the quadratic formula. Give your answer in surd form.
10
RSolve x2 + 4x + 1 = 0 using the quadratic formula. Give your answer in surd form.
4.5

Forming and solving quadratic equations

Year 9
Form and solve a quadratic equation from a context
Video lesson: Forming and solving quadratic equationsComing soon
A rectangle is 3 cm longer than it is wide, and its area is 40 cm2. Let the width be x. Then the length is (x + 3), and the area gives a quadratic equation to solve.

To form a quadratic equation, write each quantity in terms of x, use the information to make an equation, then rearrange it so that one side is zero. Solve it by factorising. Then check which solutions make sense in the context. A length or a number of pupils cannot be negative.

Key ideas

  • Let the unknown be x, and write the other quantities in terms of x.
  • Form an equation, then expand and rearrange so that one side is 0.
  • Factorise, solve, and then reject any solution that does not fit the context.
  • Check your answer in the original question, not just in the equation.
Worked example 4.5.1
A rectangle has width x cm and length (x + 3) cm. Its area is 40 cm2. Find the width.

Method. Form a quadratic from the area, then solve it. Reject any negative length.

  1. Write the sides in terms of x. Let the width be x cm. The length is x + 3 cm.
    width = x, length = x + 3
  2. Form the equation. The area is length × width, and it equals the given area.
    x(x + 3) = 40
  3. Expand and rearrange. Expand the bracket, then take the area away from both sides.
    x2 + 3x − 40 = 0
  4. Factorise. Find two numbers that multiply to -40 and add to 3: 8 and -5.
    (x − 5)(x + 8) = 0
  5. Solve each bracket. x = 5 or x = -8. A length cannot be negative, so x = 5.
    x = 5
  6. Write the answer. The width is 5 cm, and the length is 8 cm.
    5 cm

Check. 5 × 8 = 40, which is the given area.

Watch out. A negative solution is not a valid length, even though it solves the equation.

Worked example 4.5.2
Two consecutive positive whole numbers have a product of 72. Find them.

Method. Use the product to form a quadratic, solve it, then choose the positive solution.

  1. Write the two integers. Let the first integer be n. The next one is n + 1.
    n and n + 1
  2. Form the equation. Their product is the given number.
    n(n + 1) = 72
  3. Expand and rearrange. Expand the bracket and take P away from both sides.
    n2 + n − 72 = 0
  4. Factorise. The pattern (n − k)(n + k + 1) gives n2 + n − k(k + 1). Here k = 8.
    (n − 8)(n + 9) = 0
  5. Solve and reject the negative root. n = 8 or n = -9. The integers are positive, so n = 8.
    n = 8
  6. Write the answer. The two integers are the number and the next one.
    8 and 9

Check. 8 × 9 = 72.

Watch out. The negative solution gives negative integers, which the question does not ask for.

Worked example 4.5.3
A rectangle has length 2 cm more than its width, and its area is 24 cm2. Find the width.

Method. Form a quadratic from the area, then solve it. Reject any negative length.

  1. Write the sides in terms of x. Let the width be x cm. The length is x + 2 cm.
    width = x, length = x + 2
  2. Form the equation. The area is length × width, and it equals the given area.
    x(x + 2) = 24
  3. Expand and rearrange. Expand the bracket, then take the area away from both sides.
    x2 + 2x − 24 = 0
  4. Factorise. Find two numbers that multiply to -24 and add to 2: 6 and -4.
    (x − 4)(x + 6) = 0
  5. Solve each bracket. x = 4 or x = -6. A length cannot be negative, so x = 4.
    x = 4
  6. Write the answer. The width is 4 cm, and the length is 6 cm.
    4 cm

Check. 4 × 6 = 24, which is the given area.

Watch out. A negative solution is not a valid length, even though it solves the equation.

Watch out

Watch out: a negative solution to the equation may be correct algebraically but wrong in the context. Always check.

Exam tip

Exam tip. Write one line saying what x stands for, for example "Let the width be x cm".

Exercise 4E  Forming and solving quadratic equations

Non-calculator
1
PSA rectangle has length 3 cm more than its width. The width is x cm, and the area is 40 cm2. Find the width.
2
PSA rectangle has length 2 cm more than its width. The width is x cm, and the area is 24 cm2. Find the width.
3
PSA rectangle has length 1 cm more than its width. The width is x cm, and the area is 42 cm2. Find the width.
4
PSA rectangle has length 4 cm more than its width. The width is x cm, and the area is 21 cm2. Find the width.
5
PSA rectangle has length 2 cm more than its width. The width is x cm, and the area is 63 cm2. Find the width.
6
PSTwo consecutive positive whole numbers have a product of 72. Find the two numbers.
7
PSTwo consecutive positive whole numbers have a product of 110. Find the two numbers.
8
PSTwo consecutive positive whole numbers have a product of 56. Find the two numbers.
9
PSTwo consecutive positive whole numbers have a product of 90. Find the two numbers.
10
PSTwo consecutive positive whole numbers have a product of 132. Find the two numbers.

Chapter summary

SkillMethod
Expand double bracketsMultiply every term in one bracket by every term in the other, then collect like terms.
Factorise ax2 + bx + cac method: find two numbers that multiply to ac and add to b. Split, then factorise in pairs.
Difference of two squaresa2 − b2 = (a − b)(a + b).
Solve by factorisingPut in factor form, set each bracket to 0, and solve each equation.
Solve x2 + kx = 0Take out the x: x(x + k) = 0, so x = 0 or x = −k.
Completing the square (extension)Write as (x + b/2)2 − (b/2)2 + c, then take the square root.
Quadratic formula (extension)x = (−b ± √(b2 − 4ac)) / (2a).

Can you…?

  • I can expand a pair of brackets, including squared brackets
  • I can factorise ax2 + bx + c and a difference of two squares
  • I can solve a quadratic equation by factorising, and give both solutions
  • Extension: I can solve a quadratic by completing the square or the quadratic formula
  • I can form a quadratic equation from a rectangle or consecutive numbers, and reject an answer that does not fit

Key terms

quadratic
an expression or equation with x2 as its highest power
zero product property
if A × B = 0 then A = 0 or B = 0
ac method
a way of factorising ax2 + bx + c by splitting the middle term
discriminant
the part b2 − 4ac under the square root in the quadratic formula
root (solution)
a value of x that makes a quadratic equal to zero

Exercise 4Z  Check-up

Non-calculator
1
Expand and simplify (x + 4)2.
2
Expand and simplify (3x − 2)(x + 5).
3
Factorise 3x2 + 8x + 4.
4
Factorise x2 − 36.
5
Solve (x − 3)(x + 7) = 0.
6
Solve x2 + 6x = 0.
7
Solve (2x + 5)(x − 1) = 0.
8
Solve x2 − 3x − 10 = 0 by factorising.
9
Solve x2 + 2x − 2 = 0 by completing the square. Give your answer in surd form.
10
A rectangle has width x cm and length (x + 4) cm. Its area is 21 cm2. Find the width.

Total 23 marks. Allow about 23 minutes.

Exercise 4X  Practice test questions

Non-calculator
1
FExpand and simplify (2x + 3)(x − 1).[2 marks]
2
FFactorise 3x2 − 5x − 2.[2 marks]
3
FFactorise 4x2 − 25.[1 mark]
4
FSolve (2x + 1)(x − 4) = 0.[2 marks]
5
FSolve x2 − 5x = 0.[2 marks]
6
PSSolve x2 − 4x − 3 = 0 by completing the square. Give your answer in surd form.[3 marks]
7
PSSolve x2 − 5x + 2 = 0 using the quadratic formula. Give your answer in surd form.[3 marks]
8
PSA rectangle has width x cm and length (x + 6) cm. Its area is 16 cm2. Find the width.[3 marks]
9
PSTwo consecutive positive whole numbers have a product of 132. Find them.[3 marks]
10
RAli says that (x + 3)2 = x2 + 9. Explain his mistake, and expand (x + 3)2 correctly.[2 marks]
← Chapter 3: Recurring decimals, percentages and growthChapter 5: Simultaneous equations and inequalities →

Websink Prep UK is an independent site. It is not affiliated with, endorsed by or sponsored by AQA, Pearson Edexcel, OCR, WJEC, Eduqas, CCEA, SQA, GL Assessment, CEM or the Department for Education. All exam board names and trademarks belong to their owners. Past papers are linked from the exam boards' own websites and are not hosted here.

About Websink Prep UK · Find a tutor · Privacy · Terms · Cookie settings