A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

You can solve quadratic equations of the form $ax^2 + bx + c = 0$ through factorising, completing the square or using the quadratic formula.

A Level   The quickest and easiest way to solve quadratic equations is by factorising.

You need to be able to spot ‘disguised‘ quadratics involving a function of $x$, $f(x)$, instead of $x$ itself. You need to use the substitution $y=f(x)$ and solve for $y$, and then use these to find the values of $x$.

Example: Solve $x^2 + 2x = 15$ through factorising.

Rearrange the equation into the form $ax^2 + bx + c = 0$:

$x^2 + 2x - 15 = 0$

Then, solve the equation by factorising:

$(x-3)(x+5) = 0$

So,

$x - 3 = 0 \Rightarrow \textcolor{blue}{x = 3}$  and  $x + 5 = 0 \Rightarrow \textcolor{blue}{x = -5}$

A Level   ## Completing the Square Method

The following method can be used to complete the square of a quadratic expression:

Step 1: Rearrange the quadratic in the form

$ax^2 + bx + c$

Step 2: Take out a factor of $a$ out of the $x^2$ and $x$ terms:

$a \left( x^2 + \dfrac{b}{a} x \right) + c$

Step 3: Halve the coefficient of $x$ and rewrite the brackets as one squared bracket:

$a \left(x + \textcolor{limegreen}{\dfrac{b}{2a}} \right)^2$

Step 4: Add $\textcolor{red}{d}$ to the bracket to complete the square:

$a \left(x + \textcolor{limegreen}{\dfrac{b}{2a}} \right)^2 + \textcolor{red}{d}$

Step 5: Find $\textcolor{red}{d}$ by setting this expression equal to the original expression:

$a \left(x + \textcolor{limegreen}{\dfrac{b}{2a}} \right)^2 + \textcolor{red}{d} = ax^2 + bx + c$

Solving this gives

$\textcolor{red}{d} = \left( c - \dfrac{b^2}{4a} \right)$

Step 6: Put it all together:

$a \left(x + \dfrac{b}{2a} \right)^2 + \left( c - \dfrac{b^2}{4a} \right) = ax^2 + bx + c$

A Level   ## Completing the Square Formula

You do not need to do the full steps of working from above when complete the square of a quadratic, you can just express it in the form,

$ax^2 + b x + c = a \left(x + \textcolor{limegreen}{e} \right)^2 + \textcolor{red}{d}$

where

$\textcolor{limegreen}{e} =\dfrac{b}{2a} \,\,\,$ and $\,\, \textcolor{red}{d} = c-\dfrac{b^2}{4a} = c-a \textcolor{limegreen}{e}^2$

A Level   ## Solving Quadratics through Completing the Square

Completing the square isn’t the easiest method to solve quadratics, but it is useful when finding exact solutions – i.e. solutions involving surds etc.

Example: Find the exact solutions to $2x^2 - 12x + 14 = 0$ by completing the square.

So $a = 2$, $b = -12$ and $c = 14$.

Hence,

\begin{aligned} \textcolor{limegreen}{d} &= \dfrac{b}{2a} = \dfrac{-12}{2 \times 2} = \textcolor{limegreen}{-3} \\[1.2em] \textcolor{red}{e} &= c - ad^2 = 14 - 2 (-3)^2 = 14 - 18 = \textcolor{red}{-4} \end{aligned}

So,

$2x^2 - 12x + 14 = 2(x \textcolor{limegreen}{-3})^2 \textcolor{red}{- 4}$

Put this equal to $0$ and solve for $x$:

\begin{aligned} 2(x-3)^2 - 4 &= 0 \\ 2(x-3)^2 &= 4 \\ (x-3)^2 &= 2 \\ x - 3 &= \pm \sqrt{2} \end{aligned}

Hence,

$\textcolor{blue}{x = 3 + \sqrt{2}}$  and  $\textcolor{blue}{x = 3 - \sqrt{2}}$

A Level   ## Note:

To solve some equations it may be easier to use the quadratic formula instead of factorising or completing the square, e.g. when the values of $a$, $b$ and $c$ are large. However the questions usually won’t tell you which method to use.

A Level   Solve $x^4 - 13x^2 + 36 = 0$

[3 marks]

$f(x) = x^2$, so let $y = x^2$ and substitute into the equation:

$y^2 - 13y + 36 = 0$

using $x^4 = (x^2)^2 = y^2$

Then, solve the quadratic by factorising:

$(y-9)(y-4) = 0$

$y - 9 = 0 \Rightarrow \textcolor{purple}{y = 9}$  and  $y-4=0 \Rightarrow \textcolor{purple}{y = 4}$

Use these $y$ values to find the values of $x$:

$y = 9 \Rightarrow x^2 = 9 \Rightarrow x = \sqrt{9} \Rightarrow \textcolor{blue}{x = \pm 3}$

$y = 4 \Rightarrow x^2 = 4 \Rightarrow x = \sqrt{4} \Rightarrow \textcolor{blue}{x = \pm 2}$

A Level   ## Example Questions

Rearrange the equation into the form $ax^2 + bx + c = 0$:

$2x^2 - 7x + 3 = 0$

Then, solve the equation through factorising:

$(2x-1)(x-3) = 0$

So,

$2x-1 = 0 \Rightarrow x = \dfrac{1}{2}$  and  $x-3 = 0 \Rightarrow x = 3$

$f(x) = x^{\frac{1}{2}}$ so let $y = x^{\frac{1}{2}}$ and substitute into the equation:

$y^2 - y - 6 = 0$

using $x = (x^{\frac{1}{2}})^2 = y^2$

Then, solve the quadratic by factorising:

$(y-3)(y+2) = 0$

$y - 3 = 0 \Rightarrow y = 3$  and  $y+2 = 0 \Rightarrow y = -2$

Use these values of $y$ to find the values of $x$:

$y = 3 \Rightarrow x^{\frac{1}{2}} = 3 \Rightarrow x = 3^2 = 9$

$y = -2 \Rightarrow x^{\frac{1}{2}} = -2 \Rightarrow x = (-2)^2 = 4$

a) $a = 3$, $b = -30$ and $c = 73$.

Hence,

$d = \dfrac{b}{2a} = \dfrac{-30}{2 \times 3} = -5$

and

$e = c - ad^2 = 73 - 3(-5)^2 = 73 - 75 = -2$

So,

$3x^2 - 30x + 73 = 3(x-5)^2 - 2$

b) $3(x-5)^2 - 2 = 0$

So,

$3(x-5)^2 = 2$

$(x-5)^2 = \dfrac{2}{3}$

$x - 5 = \pm \sqrt{\dfrac{2}{3}} = \dfrac{\sqrt{6}}{3}$

$x = 5 + \dfrac{\sqrt{6}}{3} = 5.82$ ($2$ dp)

$x = 5 - \dfrac{\sqrt{6}}{3} = 4.18$ ($2$ dp)

A Level

A Level

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