# Completing the Square

## Completing the Square Revision

**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

- To help us solve the quadratic equation.
- 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.

**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}

**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

**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}}

## Completing the Square Example Questions

**Question 1:** Write m^2 + 5m + 6 in the form (x+a)^2+b, where a and b are constants to be determined.

**[3 marks]**

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}

**Question 2:** Write 2x^2 - 8x + 10 in the form a(x + b)^2 + c, where a, b, and c are constants to be determined.

**[4 marks]**

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.

**Question 3:** Write x^2 - 2mx + n in the form a(x + b)^2 + c, where a, b, and c are constants to be determined.

**[3 marks]**

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)

**Question 4:** Use completing the square to find the exact solutions of, z^2 + 14z - 1 = 0

**[3 marks]**

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}

**Question 5:** Use completing the square to find the exact solutions of, 4x= 3-x^2

**[5 marks]**

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}