Solving Quadratic Equations (Edexcel IGCSE Further Maths)

Spec point — 2B

Solving Quadratic Equations

You should be familiar with solving quadratic equations from your IGCSE Mathematics course. This is a quick revision guide about the different methods and when to use them.

When should I solve by factorisation?

• When the question asks to solve by factorisation

  • For example, part (a) Factorise $6x^{2}+7x-3$, part (b) Solve $6x^{2}+7x-3=0$

    • Factorises as $( 3 x - 1 ) ( 2 x + 3 ) = 0$
    • Solutions are $x=\frac{1}{3}$ and $x=-\frac{3}{2}$

• When solving two-term quadratic equations

  • For example, solve $x^{2}-4x=0$

    • Take out a common factor of $x$ to get $x ( x - 4 ) = 0$
    • Solutions are $x=0$ and $x=4$

  • For example, solve $x^{2}-9=0$

    • Use difference of two squares to factorise it as $( x + 3 ) ( x - 3 ) = 0$
    • Solutions are $x=-3$ and $x=3$
    • (Could also rearrange to $x^{2}=9$ and use ±√ to get $x=\pm3$)

When should I use the quadratic formula?

• When the question says to leave solutions correct to a given accuracy (2 decimal places, 3 significant figures etc)

• When the quadratic formula may be faster than factorising

  • It's quicker to solve $36x^{2}+33x-20=0$ using the quadratic formula than by factorisation

• If in doubt, use the quadratic formula - it always works

• You must remember the formula however - it isn't on the exam formula sheet

• If $ax^{2}+bx+c=0$, the solutions are

  • $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

When should I solve by 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

• When making $x$ the subject of harder formulae containing $x^{2}$ and $x$ terms

  • For example, make $x$ the subject of the formula $x^{2}+6x=y$

    • 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: $x=-3\pm\sqrt{y+9}$

• Like the quadratic formula, completing the square will always work

  • But it is not always quick or easy to use the method

Exam Hint

• Some calculators can solve quadratic equations

  • Even if you need to show working you can use a calculator to check your solutions
  • If the calculator solutions are whole numbers or fractions (with no square roots), this means the quadratic can be factorised

Worked Example (a) Solve $x^{2}-7x+2=0$, 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, putting 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

$x=6.7015...\mathrm{or}x=0.2984...$* *

Round your final answers to 2 decimal places

$x=6.70\mathrm{or}x=0.30$ (2 d.p.)

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

Method 1 If you cannot spot the factorisation, use the quadratic formula Substitute* $a=16$,* $b=-82$ and $c=45$ into the formula, putting 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

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

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

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

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

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

Write $x^{2}+6x$ as $( x + p )^{2} - p^{2}$ ^{* *}

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

Replace $x^{2}+6x$* **with *$( x + 3 )^{2} - 9$ in the equation

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

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

$x=-5\mathrm{or}x=-1$

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