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O-Level Mathematics · Secondary 3 and 4

Quadratic Equations

A quadratic equation has the form ax² + bx + c = 0, where a ≠ 0. In O-Level Mathematics you solve quadratics by factorising, by completing the square and by using the quadratic formula, and you link the solutions to the x-intercepts of the graph y = ax² + bx + c. Quadratics appear inside many other questions, from word problems about area and speed to graphs and coordinate geometry, so being quick and accurate here saves time across the whole paper.

What the syllabus expects

  • Solve quadratic equations by factorisation
  • Solve quadratic equations by completing the square
  • Solve quadratic equations using the formula x = (−b ± √(b² − 4ac)) / (2a), giving answers to the required accuracy
  • Form quadratic equations from word problems, solve them and reject answers that do not make sense
  • Relate the solutions of ax² + bx + c = 0 to the x-intercepts of the graph y = ax² + bx + c
  • Use the completed-square form to find the turning point and line of symmetry of a quadratic graph

Factorise first

Always rearrange so that one side is zero before you do anything else. If the expression factorises, the method is fast: x² − 5x − 14 = 0 becomes (x − 7)(x + 2) = 0. A product is zero only when one of its factors is zero, so x − 7 = 0 or x + 2 = 0, which gives x = 7 or x = −2. This zero-product idea is the reason the right-hand side must be 0. If you had (x − 7)(x + 2) = 4, you could not conclude anything from the separate brackets. Look out for special forms. A difference of two squares such as 4x² − 9 = 0 factorises to (2x − 3)(2x + 3) = 0, so x = 3/2 or x = −3/2. An equation with no constant term, such as 3x² = 12x, should be written as 3x² − 12x = 0 and factorised to 3x(x − 4) = 0, giving x = 0 or x = 4. Never divide both sides by x, because that throws away the solution x = 0.

Completing the square

Completing the square rewrites x² + bx + c as (x + b/2)² − (b/2)² + c. For example, x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7. To solve x² + 6x + 2 = 0, write (x + 3)² = 7, take the square root of both sides to get x + 3 = ±√7, and so x = −3 + √7 or x = −3 − √7, which is x = −0.354 or x = −5.65 to 3 significant figures. Remember the ± sign, since both a positive and a negative number square to 7. If the coefficient of x² is not 1, divide the whole equation by it first. The completed-square form also describes the graph. Since (x + 3)² is never negative, y = (x + 3)² − 7 has a minimum value of −7 when x = −3, so the minimum point is (−3, −7) and the line of symmetry is x = −3.

The quadratic formula

The formula works for every quadratic, including those that do not factorise: x = (−b ± √(b² − 4ac)) / (2a). Write down a, b and c with their signs before substituting. For 2x² − 3x − 7 = 0, a = 2, b = −3 and c = −7, so b² − 4ac = 9 + 56 = 65 and x = (3 ± √65) / 4, giving x = 2.77 or x = −1.27 to 2 decimal places. Put the whole numerator over 2a, not just the square root. If b² − 4ac turns out to be negative, its square root cannot be found, so the equation has no real solutions, which matches a graph that does not cross the x-axis. When a question asks for answers to 2 decimal places or 3 significant figures, it is a strong hint that the formula or completing the square is expected, because the equation is unlikely to factorise.

Word problems

Many questions ask you to form the equation yourself. Let the unknown be a clear quantity with units, such as 'let the width be x cm', then translate each sentence. If a rectangle is 3 cm longer than it is wide and has an area of 54 cm², then x(x + 3) = 54, so x² + 3x − 54 = 0, which factorises to (x + 9)(x − 6) = 0. The solution x = −9 is rejected because a width cannot be negative, so the width is 6 cm and the length is 9 cm. Speed and time questions often lead to fractions, such as 120/x − 120/(x + 10) = 1. Multiply every term by x(x + 10) to clear the denominators, which gives 120(x + 10) − 120x = x(x + 10), then simplify to x² + 10x − 1200 = 0 and solve to get x = 30 or x = −40. Always state which solution you reject and why, and answer the question in a sentence with units.

Worked example

Solve the equation (2x + 1)(x − 3) = 7, giving your answers correct to 2 decimal places.

  1. Expand the left-hand side: 2x² − 6x + x − 3 = 2x² − 5x − 3.
  2. Make one side zero: 2x² − 5x − 3 − 7 = 0, so 2x² − 5x − 10 = 0.
  3. Identify a = 2, b = −5 and c = −10.
  4. Work out b² − 4ac = (−5)² − 4(2)(−10) = 25 + 80 = 105.
  5. Substitute into the formula: x = (5 ± √105) / 4, where √105 = 10.247 to 3 decimal places.
  6. So x = (5 + 10.247) / 4 = 3.8117 or x = (5 − 10.247) / 4 = −1.3117.
  7. Check by substitution: 2(3.8117)² − 5(3.8117) − 10 is approximately 0, as required.

Answer: x = 3.81 or x = −1.31, correct to 2 decimal places

Common mistakes

  • Splitting a product that is not equal to zero. From (x − 2)(x + 1) = 4 you cannot write x − 2 = 4 or x + 1 = 4. Expand, collect everything on one side to get x² − x − 6 = 0, and factorise again to get x = 3 or x = −2.
  • Dividing both sides by x. Turning x² = 5x into x = 5 loses the solution x = 0. Rearrange to x² − 5x = 0 and factorise to x(x − 5) = 0.
  • Sign errors with b in the formula. If b = −5, then −b = 5 and b² = 25, not −25. Writing a, b and c down with their signs before substituting prevents most of these slips.
  • Forgetting the ± when square-rooting. (x + 3)² = 7 gives x + 3 = √7 or x + 3 = −√7, so there are two answers, not one.
  • Rounding too early. Keep √105 as 10.247 or better until the last line, then round to the accuracy asked for. Rounding √105 to 10 first would give 3.75 instead of 3.81.
  • Leaving in an impossible answer. In a word problem, a negative length, a negative speed or a fractional number of people must be rejected with a short reason.

Exam tips

  • If the question says give your answers to 2 decimal places, go straight to the formula rather than spending time hunting for factors that do not exist.
  • Write the formula out in full once, then substitute using brackets, such as (−5)² and 4(2)(−10). This makes your working easy to follow if you need to find a slip.
  • In word problems, show the step where you reject a solution and give the reason in a few words. The final answer should be a sentence with units.
  • Check solutions by substituting back into the original equation, not the rearranged one, so that you also catch errors in your expansion.

Practice questions

1. Solve x² − 7x + 12 = 0.

Show answer

x = 3 or x = 4. Factorise to (x − 3)(x − 4) = 0, since (−3) × (−4) = 12 and (−3) + (−4) = −7, then set each bracket equal to zero.

2. A rectangle has a length that is 3 cm more than its width. Its area is 40 cm². Find its width and its length.

Show answer

Width 5 cm, length 8 cm. Let the width be x cm, so x(x + 3) = 40, giving x² + 3x − 40 = 0 and (x + 8)(x − 5) = 0. Reject x = −8 because a width cannot be negative, so x = 5 and the length is 5 + 3 = 8 cm.

3. Express x² − 8x + 5 in the form (x − a)² + b. Hence write down the minimum point of y = x² − 8x + 5 and solve x² − 8x + 5 = 0, giving your answers to 2 decimal places.

Show answer

x² − 8x + 5 = (x − 4)² − 16 + 5 = (x − 4)² − 11, so the minimum point is (4, −11). Solving (x − 4)² = 11 gives x − 4 = ±√11 = ±3.3166, so x = 7.32 or x = 0.68.

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