Deciding the Quadratic Method (Edexcel IGCSE Maths A (Modular))

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Mark Curtis

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Maths

Deciding the Quadratic Method

If you have to solve a quadratic equation but are not told which method to use, here is a guide for what to do.

When should I solve by factorisation?

Use factorisation when the question asks to solve by factorisation
- For example
  - part (a) Factorise 6x² + 7x – 3
  - part (b) Solve 6x² + 7x – 3 = 0
Use factorisation when solving two-term quadratic equations
- For example, solve x² – 4x = 0
  - Take out a common factor of x to get x(x – 4) = 0
  - So x = 0 and x = 4
- For example, solve x² – 9 = 0
  - Use the difference of two squares to factorise it as (x + 3)(x – 3) = 0
  - So x = -3 and x = 3
  - (Or rearrange to x² = 9 and use ±√ to get x = ±3)
Factorising can often be the quickest way to solve a quadratic equation

When should I use the quadratic formula?

Use the quadratic formula when the question says to leave solutions correct to a given accuracy (2 decimal places, 3 significant figures etc)
- This is a hint that the equation will not factorise
Use the quadratic formula when it may be faster than factorising
- It's quicker to solve 36x² + 33x – 20 = 0 using the quadratic formula than by factorisation
Use the quadratic formula if in doubt, as it always works

When should I solve by completing the square?

Use completing the square when part (a) of a question says to complete the square and part (b) says to use part (a) to solve the equation
Use completing the square when making x the subject of harder formulae containing both x² and x terms
- For example, make x the subject of the formula x² + 6x = y
  - Complete the square: (x + 3)² – 9 = y
  - Add 9 to both sides: (x + 3)² = y + 9
  - Take square roots and use ±: $x + 3 = \pm \sqrt{y + 9}$
  - Subtract 3: $x = - 3 \pm \sqrt{y + 9}$
Completing the square always works
- But it's not always quick or easy to do

Exam Tip

If your calculator solves quadratic equations, use it to check your solutions
If the solutions on your calculator are whole numbers or fractions (with no square roots), this means the quadratic equation does factorise

Worked Example

(a) Solve $x^{2} - 7 x + 2 = 0$ , giving your answers correct to 2 decimal places.

“Correct to 2 decimal places” suggests using the quadratic formula
Substitute a = 1, b = -7 and c = 2 into the formula
Put brackets around any negative numbers

$x = \frac{- (- 7) \pm \sqrt{{(- 7)}^{2} - 4 \times 1 \times 2}}{2 \times 1}$

Use a calculator to find each solution

x = 6.70156… or 0.2984...

Round your final answers to 2 decimal places

x = 6.70 or x = 0.30 (2 d.p.)

(b) Solve $16 x^{2} - 82 x + 45 = 0$ .

Method 1
If you cannot spot the factorisation, use the quadratic formula
Substitute a = 16, b = -82 and c = 45 into the formula
Put brackets around any negative numbers

$x = \frac{- (- 82) \pm \sqrt{{(- 82)}^{2} - 4 \times 16 \times 45}}{2 \times 16}$

Use a calculator to find each solution

x = $\frac{9}{2}$ or x = $\frac{5}{8}$

Method 2
If you do spot the factorisation, (2x – 9)(8x – 5), then use that method instead

$(2 x - 9) (8 x - 5) = 0$

Set the first bracket equal to zero

$2 x - 9 = 0$

Add 9 to both sides then divide by 2

$\begin{array}{rcl} 2 x & = & 9 \\ x & = & \frac{9}{2} \end{array}$

Set the second bracket equal to zero

$8 x - 5 = 0$

Add 5 to both sides then divide by 8

$\begin{array}{rcl} 8 x & = & 5 \\ x & = & \frac{5}{8} \end{array}$

x = $\frac{9}{2}$ or x = $\frac{5}{8}$

This question wants you to complete the square first
Find p (by halving the middle number)

$p = \frac{6}{2} = 3$

Write x² + 6x as (x + p)² - p²

$\begin{array}{rcl} x^{2} + 6 x & = & {(x + 3)}^{2} - 3^{2} \\ = & {(x + 3)}^{2} - 9 \end{array}$

Replace x² + 6x with (x + 3)² – 9 in the equation

$\begin{array}{rcl} {(x + 3)}^{2} - 9 + 5 & = & 0 \\ {(x + 3)}^{2} - 4 & = & 0 \end{array}$

Now solve it
Make x the subject of the equation (start by adding 4 to both sides)

${(x + 3)}^{2} = 4$

Take square roots of both sides (include a ± sign to get both solutions)

$x + 3 = \pm \sqrt{4} = \pm 2$

Subtract 3 from both sides

$x = - 3 \pm 2$

Find each solution separately using + first, then - second

x = - 1 or x = - 5

Even though the quadratic factorises to (x + 5)(x + 1), this is not the method asked for in the question

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