Completing the Square Questions, Revision and Worksheets

GCSE 6 - 7GCSE 8 - 9AQAEdexcelOCRWJECHigherAQA 2022OCR 2022WJEC 2022

Completing the Square

Completing the square is a method of changing the way that a quadratic is expressed. There are two reasons we might want to do this, and they are

1. To help us solve the quadratic equation.
2. To find the coordinates of the minimum (or maximum) point of a quadratic graph.

Make sure you are happy with the following topics before continuing.

Level 6-7

Completing the Square Formula

What does “completing the square” mean? Well, it involves taking a quadratic equation, and expressing it in the form,

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

where

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

or

$\textcolor{blue}{e} \textcolor{black}{=c-a}\textcolor{red}{d}\textcolor{black}{^2}$

Level 8-9
Level 6-7

Skill 1: Completing the Square $a=1$

Solving quadratics via completing the square can be tricky, first we need to write the quadratic in the form $(x+\textcolor{red}{d})^2 + \textcolor{blue}{e}$ then we can solve it. Since $a=1$, this can be done in $4$ easy steps.

Example: By completing the square, solve the following quadratic $x^2+6x +3=1$

Step 1: Rearrange the equation so it is $=0$

\begin{aligned}(-1)\,\,\,\,\,\,\,\,\,x^2+6x+3 &=1 \\ x^2 +6x +2&=0\end{aligned}

Step 2: Half the coefficient of $x$, so in this case $\textcolor{red}{d}=6\div 2=\textcolor{red}{3}$, and add it in the place of $\textcolor{red}{d}$

$(x+\textcolor{red}{3})^2 + \textcolor{blue}{e}$

Step 3: Next we need to find $\textcolor{blue}{e}$ which equals the constant at the end of the quadratic, $+2$, minus $\textcolor{red}{d^2}$, then replace $\textcolor{blue}{e}$ in the equation ($\textcolor{blue}{e} \textcolor{black}{=c-}\textcolor{red}{d}\textcolor{black}{^2}$ as $\textcolor{black}{a=1}$).

\begin{aligned}\textcolor{blue}{e} &= 2 -\textcolor{maroon}{9} \\ \textcolor{blue}{e} &= -7\end{aligned}

$(x+\textcolor{red}{3})^2 \textcolor{blue}{-7} = 0$

Step 4: Now we have the equation in this form we can solve the equation.

\begin{aligned}(+7)\,\,\,\,\,\,\,\,\,(x+3)^2 -7 &= 0 \\ (\sqrt{})\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(x+3)^2 &= 7 \\ (-3)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, x+3 &= \pm \sqrt{7} \\ x &= \pm \sqrt{7}- 3\end{aligned}

This gives the solutions to be

$\sqrt{7} - 3$ and $-\sqrt{7} - 3$

Remember: A square root can have both a positive and negative solution

Level 8-9
Level 8-9

Skill 2: Complete the square $a>1$

When $a\neq1$ things become a little trickier. The majority of the method is the same but with an additional factorisation step at the beginning.

Example: Write $3x^2 + 5x-3$ in the form $\textcolor{limegreen}{a}(x+\textcolor{red}{d})^2+\textcolor{blue}{e}$

Step 1: Factorise the first two terms by the coefficient in front of $x^2$, this now becomes $\textcolor{limegreen}{a}$

$\textcolor{limegreen}{3}\bigg(x^2 + \dfrac{5}{3}x \bigg)-3$

Step 2: Half the coefficient of $x$ and write it in the place of $\textcolor{red}{d}$

$\dfrac{5}{3} \div 2 = \textcolor{red}{\dfrac{5}{6}}$

$\textcolor{limegreen}{3}\bigg(x+\textcolor{red}{\dfrac{5}{6}}\bigg)^2+\textcolor{blue}{e}$

Step 3: Next we need to find $\textcolor{blue}{e}$ which equals the constant at the end of the quadratic, $-3$, minus the ‘non-$x$‘ result from expanding the brackets. ($\textcolor{blue}{e} \textcolor{black}{=c-a}\textcolor{red}{d}\textcolor{black}{^2}$)

$3\bigg(x+\dfrac{5}{6}\bigg)^2 = 3\bigg(x+\dfrac{5}{6}\bigg)\bigg(x + \dfrac{5}{6}\bigg) = 3x^2 +5x + \dfrac{25}{12}$

$3\bigg(x+\dfrac{5}{6}\bigg)^2 - 3 - \dfrac{25}{12}$

$= 3\bigg(x+\dfrac{5}{6}\bigg)^2 -\dfrac{61}{12}$

So the completed square is

$\textcolor{limegreen}{3}\bigg(x+\textcolor{red}{\dfrac{5}{6}}\bigg)^2 \textcolor{blue}{-\dfrac{61}{12}}$

Level 8-9

Example Questions

The coefficient of $m$ term is $5$, and half of $5$ is $\frac{5}{2}$, so we get

$m^2+5m+6=\left(m+\dfrac{5}{2}\right)^2 + 6 - \left(\dfrac{5}{2}\right)^2$

Considering the value of $b$,

$6-\left(\dfrac{5}{2}\right)^2 = 6 - \dfrac{25}{4}=\dfrac{24}{4}-\dfrac{25}{4}=-\dfrac{1}{4}$

Now we know the constant that goes outside the bracket, the final result of completing the square is,

$\left(m+\dfrac{5}{2}\right)^2-\dfrac{1}{4}$

In order to be able to apply our normal process of completing the square, we need to take a factor of $2$ out of this whole expression:

$2x^2 - 8x + 10 = 2(x^2 - 4x + 5)$

Now it looks more familiar, the coefficient of the $x$ term is $-4$, half of which is $-2$, so we get:

$2(x^2 - 4x + 5) = 2\left[(x - 2)^2 + 5 - (-2)^2\right] = 2[(x - 2)^2 + 1]$

Multiplying through by $2$ we find,

$2\left[(x -2)^2 + 1\right] = 2(x - 2)^2 + 2$

which is in the form asked for in the question.

The coefficient of $x$ term is $-2m$, and half of $-2m$ is $-m$, so we get

$x^2-2mx+n=(x-m)^2 + n - (-m)^2$

This simplifies so that the final result of completing the square is,

$(x-m)^2 + (n - m^2)$

The coefficient of the $z$ term is $14$, half of which is $7$, so we get:

$z^2 + 14z - 1 = (z + 7)^2 - 1 - 7^2 = (z + 7)^2 - 50$

Meaning our equation is now

$(z + 7)^2 - 50 = 0$

Now we must rearrange this equation to make $z$ the subject,

\begin{aligned} (z + 7)^2 &= 50 \\ z + 7 & = \pm \sqrt{50} \\ z &= -7 \pm \sqrt{50}\end{aligned}

So, the two solutions are,

$x = -7 + \sqrt{50}$ and $x = -7 - \sqrt{50}$

We have to start by writing the equation in a more familiar form,

$3-4x-x^2=0$

In this case, we have $a=-1$ so now completing the square we get,

$3-4x-x^2 = -(x+2)^2 +3+4$

Now to solve this quadratic we must rearrange it to make $x$ the subject,

\begin{aligned} -(x+2)^2 +7 &= 0 \\ -(x+2)^2 &= -7 \\ (x+2)^2 &= 7 \\ x+2 & = \pm \sqrt{7} \\ x&= -2 \pm \sqrt{7}\end{aligned}

So, the two solutions are,

$x = -2 + \sqrt{7}$ and $x = -2 - \sqrt{7}$

Related Topics

Level 4-5GCSEKS3

Level 4-5GCSEKS3

Surds

Level 6-7Level 8-9GCSE

Rearranging Formulae

Level 4-5Level 6-7GCSEKS3

Level 6-7

Level 8-9

Level 6-7

Level 8-9

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