Quadratic equations

The method

1. Rearrange the equation so that one side is zero.
2. Take out any common factor before you factorise.
3. Factorise the quadratic and set each bracket equal to zero.
4. If it will not factorise, use the quadratic formula with a, b and c.
5. Check each root in the original equation and keep the negative ones.

Mistakes that lose marks

• Losing the x = 0 root when you divide both sides by x instead of factorising.
• Factorising before moving every term to one side, so the product is not equal to zero.
• Using the formula with the wrong sign for a negative b or c.
• Giving only the positive root when the equation has two.
• Rounding a surd answer when an exact answer is required.

7 exam-style questions

These are our own questions written in the style of the exam, not past-paper questions. The mark codes are M = method, A = accuracy and B = independent mark. Try each one, then open the solution.

1. Exam-style core Solve x² − 7x + 12 = 0. [2 marks]
   Show full solution

   1. Find two numbers with product 12 and sum −7: they are −3 and −4, so (x − 3)(x − 4) = 0 M1
   2. x − 3 = 0 or x − 4 = 0, giving x = 3 or x = 4 A1

   Answer: x = 3 or x = 4

2. Exam-style core Solve 3x² − 12x = 0. [3 marks]
   Show full solution

   1. Take out the common factor: 3x(x − 4) = 0 M1
   2. 3x = 0 gives x = 0 M1
   3. x − 4 = 0 gives x = 4, so the two roots are 0 and 4 A1

   Answer: x = 0 or x = 4

3. Exam-style core Solve x² = 5x + 24. [3 marks]
   Show full solution

   1. Rearrange so one side is zero: x² − 5x − 24 = 0 M1
   2. Factorise: (x − 8)(x + 3) = 0 M1
   3. x = 8 or x = −3 A1

   Answer: x = 8 or x = −3

4. Exam-style extended Solve 2x² + 4x − 1 = 0, giving your answers in surd form. [3 marks]
   Show full solution

   1. Use x = (−b ± √(b² − 4ac)) ÷ 2a with a = 2, b = 4 and c = −1: x = (−4 ± √24) ÷ 4 M1
   2. Simplify the surd: √24 = 2√6, so x = (−4 ± 2√6) ÷ 4 M1
   3. Divide every term by 2: x = −1 ± (√6 ÷ 2) A1

   Answer: x = −1 ± (√6 ÷ 2)

5. Exam-style extended A ball is kicked and its height above the ground after t seconds is h = 12t − 5t² metres. Find the two times when the ball is 4 m above the ground. [3 marks]
   Show full solution

   1. Set 12t − 5t² = 4 and rearrange: 5t² − 12t + 4 = 0 M1
   2. Factorise: (5t − 2)(t − 2) = 0 M1
   3. t = 2/5 or t = 2 A1

   Answer: t = 2/5 s and t = 2 s

6. Exam-style extended A rectangle is 5 cm longer than it is wide. Its area is 84 cm². Find the dimensions of the rectangle. [4 marks]
   Show full solution

   1. Let the width be x cm, so the length is x + 5 cm and x(x + 5) = 84 M1
   2. Expand and factorise: x² + 5x − 84 = 0, so (x + 12)(x − 7) = 0 M1
   3. x = 7, because x = −12 is impossible for a length A1
   4. The rectangle is 7 cm by 12 cm A1

   Answer: 7 cm by 12 cm

7. Exam-style higher Solve 4x² − 12x + 7 = 0, giving your answers in the form (a ± √b) ÷ c. [4 marks]
   Show full solution

   1. Use the formula with a = 4, b = −12 and c = 7: x = (12 ± √((−12)² − 4 × 4 × 7)) ÷ 8 M1
   2. The discriminant is 144 − 112 = 32, and √32 = 4√2, so x = (12 ± 4√2) ÷ 8 M1
   3. Divide every term by 4: x = (3 + √2) ÷ 2 A1
   4. The second root is x = (3 − √2) ÷ 2 A1

   Answer: x = (3 ± √2) ÷ 2

Where this is in the syllabus

Quadratic equations appears in Cambridge IGCSE Mathematics 0580 and Pearson Edexcel International GCSE Mathematics A 4MA1. For the official specifications and past papers, use the board's own website: Cambridge IGCSE Mathematics 0580 and Pearson Edexcel International GCSE Mathematics A 4MA1.

Questions students ask