Deciding the Quadratic Solution Method (Edexcel IGCSE Maths)

Quadratic Equation Methods

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Deciding the quadratic method

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 Hint

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

Answer:

“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 $16x^{2}-82x+45=0$.

Answer:

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

Final answer: 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

$2x-9=0$

Add 9 to both sides then divide by 2

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

Set the second bracket equal to zero

$8x-5=0$

Add 5 to both sides then divide by 8

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

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

(c) By writing $x^{2}+6x+5$ in the form $( x + p )^{2} + q$, solve $x^{2}+6x+5=0$

Answer:

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

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

Write x² + 6**x **as (x + p)²* -** p*²

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

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

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

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)

$x+3=\pm\sqrt{4}=\pm2$

Subtract 3 from both sides

$x=-3\pm2$

Find each solution separately using + first, then - second

Final answer: x = - 1 or x = - 5