Quadratic Equation Methods (Cambridge (CIE) IGCSE Additional Maths): Revision Note

• Calculators can solve quadratic equations

  • Double check you've entered the equation correctly, in the correct format

  • Use this feature to check your answers where possible

  • 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
 

“Correct to 2 decimal places” suggests using the quadratic formula and/or a calculator
For accuracy, it is a good idea to use both - use the formula and calculator as normal first
Then use the quadratic solver feature to check your solutions

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 = 6.70156… or 0.2984...
 

Round your final answers to 2 decimal places

x = 6.70 or x = 0.30

If your calculator has a quadratic equation solver, use it to check your answers

(b) Solve $16x^2-82x+45=0$
 

Method 1
The coefficients are large and so the factorisation, even if possible, is hard to spot
Therefore, one method to use is the quadratic formula - it always works!
The solution below is the manual way to use a calculator, but as above, if your calculator has a quadratic solver feature, you may use that
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

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

Method 2
If you do persevere with the factorisation then use that method instead
 

$16x^2-82x+45=\left(2x-9\right)\left(8x-5\right)=0$
 

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

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

Notice this question does not use the phrase 'completing the square' but shows the form of it instead
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
 

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

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
 

$x=\pm 2-3$
 

Find each solution separately using + first, then - second

x = - 5, x = - 1

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

• While the substitution method is not compulsory, beware of skipping steps

  • e.g. it is incorrect to "jump" from the solutions of x² -3x - 4 = 0 to the solutions of (x + 5)² - 3(x + 5) - 4 = 0 by "adding 5 to them"

    • the substitution method shows you end up subtracting 5

(a) Solve  $x^8-17x^4+16=0$

This is a quadratic in x⁴ so let u = x⁴

$u^2-17u+16=0$

Solve this simpler quadratic equation, for example by factorisation

$\left(u-16\right)\left(u-1\right)=0$ 

Write out the u solutions

$u=16\mathrm{or}u=1$

Replace u with x⁴

$x^4=16\mathrm{or}{x}^4=1$

Solve these separate equations (remember an even power gives two solutions)

$x=\pm 2\mathrm{or}x=\pm 1$

Write your solutions out (it's good practice to write them in numerical order) 

$\mathit{x}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{,}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{or}\mathbf{2}$

(b) Solve  $x-\sqrt{x}-6=0$

This is a quadratic in √x so let u = √x

$u^2-u-6=0$

Solve this simpler quadratic equation, for example by factorisation

$$\begin{gathered} \left(u-3\right)\left(u+2\right)=0 \\ u=3\mathrm{or}u=-2 \end{gathered}$$

Replace u with √x and solve

$$\begin{gathered} \sqrt{x}=3\Rightarrow x=9 \\ \sqrt{x}=-2\mathrm{has}\mathrm{no}\mathrm{solutions}\mathrm{as}\sqrt{x}\ge 0 \end{gathered}$$

You can check your solutions by substituting them back into the equation If you put x = 4 as a solution by mistake then substituting will spot this error

$$\begin{gathered} 9-\sqrt{9}-6=0 \\ 4-\sqrt{4}-6=-4\ne 0 \end{gathered}$$

$\mathit{x}\mathbf{=}\mathbf{9}$