Completing the Square Questions | Worksheets and Revision | MME

# Completing the Square Questions, Revision and Worksheets

Level 8 Level 9

## What you need to know

### 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. Please see turning points of quadratic graphs.

What does “completing the square” mean? Well, it involves taking a quadratic, say

$x^2 + 4x - 12$,

and expressing it in the form

$(x + a)^2 + b$

where $a$ and $b$ are constants to be determined.

### Take Note:

So, this is how you complete the square: halve the coefficient of $x$ to find the number that goes inside the bracket, and then subtract the square of it from the constant term in your quadratic to work out what goes on the end. For the algebra-loving people amongst us, here is the completing the square formula:

$x^2 + cx + d = \left(x + \dfrac{c}{2}\right)^2 + d - \left(\dfrac{c}{2}\right)^2$

### Example 1: Completing The Square

Write $x^2+4x-12$ in the form $(x+a)^2+b$ where $a$ and $b$ are constants to be determined.

Firstly, the $a$ is found simply by halving the coefficient of $x$, so in this case $a=4\div 2=2$. Then, to find $b$, we take the constant at the end of the quadratic, -12, and subtract $a^2$ from it, so we get $b = -12 - 2^2 = -16$. Therefore, the result of completing the square is

$x^2 + 4x - 12 = (x + 2)^2 - 16$

### Example 2: Completing The Square

Write $2y^2-16y+6$ in the form $a(x+b)^2+c$, where $a, b,$ and $c$ are constants to be determined.

We need to start by taking a factor of 2 out of the whole expression to get it in the form we are use to. Doing so, we get

$2y^2-16y+6=2(y^2-8y+3)$

Now, all we need to do is complete the square on the bit inside the bracket (as we normally would) and then bring the 2 back in at the end. So, we get

$y^2-8y+3 = (y-4)^2+3-(-4)^2=(y-4)^2-13$

Remember, this is just the inside of the bracket, so if we bring the 2 back into the picture, we get

$2\left[(y-4)^2-13\right]$

Finally, expanding the square brackets (and only those), we get the final form of the expression to be

$2(y-4)^2 - 26$

This is precisely the form the question asked for.

### Example 3: Using Completing The Square to Solve a Quadratic Equation

Use completing the square to solve $x^2 + 4x + 3 = 0$.

Using the completing the square method our equation becomes

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

Now, to solve this equation for $x$ we will rearrange our equation – not expanding the brackets – to make $x$ the subject. Firstly, add 1 to both sides to get

$(x + 2)^2 = 1$

Then, square root both sides to get

$x + 2 = \pm\sqrt{1} = \pm 1$

Finally, subtracting 2 from both sides makes $x$ the subject and leaves us with

$x = \pm 1 - 2$.

Therefore, the two solutions to our quadratic are $x = -1$ and $x = -3$.

Note: Make sure you remember to include both the positive and negative square roots, or you will end up missing one of the solutions.

### 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 sqaure we get,

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

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

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