Solving Quadratic Equations (Edexcel IGCSE Further Pure Maths): Revision Note

• 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

(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
 

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

Use a calculator to find each solution 

$x=6.7015...\mathrm{or}x=0.2984...$ 

Round your final answers to 2 decimal places

$\mathit{x}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{70}\mathbf{or}\mathit{x}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{30}$  (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

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

Use a calculator to find each solution

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

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

$\left(2x-9\right)\left(8x-5\right)=0$ 

Set the first bracket equal to zero 

$2x-9=0$ 

Add 9 to both sides then divide by 2 

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

Set the second bracket equal to zero 

$8x-5=0$ 

Add 5 to both sides then divide by 8 

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

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

 (c) By writing $x^2+6x+5$ in the form ${\left(x+p\right)}^2+q$, 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 ${\left(x+p\right)}^2-p^2$
 

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

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

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

Make $x$ the subject of the equation
Start by adding 4 to both sides 

${\left(x+3\right)}^2=4$ 

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

$\mathit{x}\mathbf{=}\mathbf{-}\mathbf{5}\mathbf{or}\mathit{x}\mathbf{=}\mathbf{-}\mathbf{1}$

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