Deciding the Quadratic Solution Method (Edexcel IGCSE Maths B): Revision Note

• 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

(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 

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

Use a calculator to find each solution 

x = 6.70156… 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

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

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

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

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

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

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

(c) By writing $x^2+6x+5$ in the form ${\left(x+p\right)}^2+q$, 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 + 6x as (x + p)² - p²

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

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

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

Now solve it
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

x = - 1  or  x = - 5