Functions and algebra: Solve algebraic equations and inequalities

# Unit 2: Solve quadratic equations by completing the square

Dylan Busa

### Unit outcomes

By the end of this unit you will be able to:

• Solve a quadratic equation by completing the square.

## What you should know

Before you start this unit, make sure you can:

Here is a short self-assessment to make sure you have the skills you need to proceed with this unit.

Complete the square of the following quadratic expressions:

1. $\scriptsize {{x}^{2}}-4x-4$
2. $\scriptsize 2{{x}^{2}}-6x+1$

Solutions

1. .
\scriptsize \begin{align*}{{x}^{2}}-4x-4&={{x}^{2}}-4x+4-4-4\\&=({{x}^{2}}-4x+4)-8\\&={{(x+2)}^{2}}-8\end{align*}
2. .
\scriptsize \begin{align*} 2x^2-6x+1&=2\left(x^2-3x+\displaystyle \frac{1}{2}\right)\\ &=2\left( x^2-3x+\displaystyle \frac{9}{4}-\displaystyle \frac{9}{4}+\displaystyle \frac{1}{2}\right)\\ &=2\left( x^2-3x+\displaystyle \frac{9}{4}\right)-\displaystyle \frac{9}{2}+1\\ &=2\left(x-\displaystyle \frac{3}{2}\right)^2-\displaystyle \frac{7}{2} \end{align*}

## Introduction

As we saw in the previous unit, most of the quadratic equations that you will come across can be solved by factorising the quadratic expression when you have the equation in the standard form of $\scriptsize a{{x}^{2}}+bx+c=0$. You must always try to factorise this quadratic expression first before trying anything else.

However, there are some quadratic expressions that are hard to factorise. In these cases, you can use a technique called completing the square. You learnt why it is called completing the square in subject outcome 2.2 unit 3.

## Solve quadratic equation by completing the square

You should know how to complete the square of a quadratic expression. However, letâ€™s look at one example before we move on.

### Example 2.1

Factorise the following quadratic expression fully by first completing the square:$\scriptsize {{x}^{2}}-10x-11$

Solution

$\scriptsize {{x}^{2}}-10x+11$ cannot be easily factorised but we can complete the square.

$\scriptsize {{x}^{2}}-10x+11={{x}^{2}}-10x+25-25+11$

We add $\scriptsize 25$ to the expression because $\scriptsize {{\left( {\displaystyle \frac{{10}}{2}} \right)}^{2}}=25$.

However, we need to subtract $\scriptsize 25$ from the expression as well to keep the value of the whole expression the same.

\scriptsize \begin{align*}{{x}^{2}}-10x+11&={{x}^{2}}-10x+25-25+11\\&=(x-5)(x-5)-25+11\\&={{(x-5)}^{2}}-14\end{align*}

We have completed the square. But now we have a difference of two squares, so we can factorise further.

\scriptsize \begin{align*}{{(x-5)}^{2}}-14&=\left( {(x-5)+\sqrt{{14}}} \right)\left( {(x-5)-\sqrt{{14}}} \right)\\&=\left( {x-5+\sqrt{{14}}} \right)\left( {x-5-\sqrt{{14}}} \right)\end{align*}

Hopefully you can see how the technique of completing the square allows us to solve quadratic equations that are not easily factorised.

### Example 2.2

Solve for $\scriptsize x$: $\scriptsize {{x}^{2}}-4x=3$

Solution

As usual, we need to get the equation into standard form.
\scriptsize \begin{align*}{{x}^{2}}-4x&=3\\\therefore {{x}^{2}}-4x-3&=0\end{align*}

Now the quadratic expression is not easy to factorise, so we continue by completing the square.
\scriptsize \begin{align*}{{x}^{2}}-4x-3 & =0\\\therefore {{x}^{2}}-4x+4-4-3 & =0\\\therefore {{(x-2)}^{2}}-7 & =0\end{align*}

At this point, we can either continue to factorise the resulting difference of two squares or proceed as follows. The choice is yours.
\scriptsize \begin{align*}{{(x-2)}^{2}}-7 & =0\\ \therefore {{(x-2)}^{2}} & =7&&\text{Take the square root of both sides}\\ \therefore (x-2) & =\pm \sqrt{7}&& \text{Remember that the RHS is }\pm \sqrt{7}\\ \therefore x=2+\sqrt{7}&\text{ or }x=2-\sqrt{7}\end{align*}

Make sure that you can get the same roots by factorising the difference of two squares.

$\scriptsize x=2+\sqrt{7}$:
\scriptsize \begin{align*}\text{LHS}&={{x}^{2}}-4x\\&={{(2+\sqrt{7})}^{2}}-4\left( {2+\sqrt{7}} \right)\\&=4+4\sqrt{7}+7-8-4\sqrt{7}\\&=3\\&=\text{RHS}\end{align*}

$\scriptsize x=2-\sqrt{7}$:
\scriptsize \begin{align*}\text{LHS}&={{x}^{2}}-4x\\&={{(2-\sqrt{7})}^{2}}-4\left( {2-\sqrt{7}} \right)\\&=4-4\sqrt{7}+7-8+4\sqrt{7}\\&=3\\&=\text{RHS}\end{align*}

Both solutions are valid. Therefore, $\scriptsize x=2+\sqrt{7} \text{or}\ x=2-\sqrt{7}$.

Note: It is always best to test your solutions with the original equation.

### Example 2.3

Solve for $\scriptsize a$: $\scriptsize 3{{a}^{2}}-9a=15$

Solution

Get the equation into standard form.
\scriptsize \begin{align*}3{{a}^{2}}-9a & =15\\\therefore 3{{a}^{2}}-9a-15 & =0\end{align*}

We cannot factorise this quadratic expression easily, so we proceed by completing the square.
\scriptsize \begin{align*}3{{a}^{2}}-9a-15 & =0\\\therefore 3({{a}^{2}}-3a-5) & =0\\ \therefore {{a}^{2}}-3a-5 & =0\\ \therefore {{a}^{2}}-3a+\displaystyle \frac{9}{4}-\displaystyle \frac{9}{4}-5 & =0\\ \therefore {{\left( {a-\displaystyle \frac{3}{2}} \right)}^{2}}-\displaystyle \frac{{29}}{4} & =0\\ \therefore {{\left( {a-\displaystyle \frac{3}{2}} \right)}^{2}} & =\displaystyle \frac{{29}}{4}&&\text{Take the square root of both sides}\\\therefore a-\displaystyle \frac{3}{2} & =\displaystyle \frac{{\pm \sqrt{{29}}}}{2}&&\text{Remember that the RHS is }\displaystyle \frac{{\pm \sqrt{{29}}}}{2}\\ \therefore a=\displaystyle \frac{3}{2}+\displaystyle \frac{{\sqrt{{29}}}}{2}&\text{ or }a=\displaystyle \frac{3}{2}-\displaystyle \frac{{\sqrt{{29}}}}{2}\\\therefore a=\displaystyle \frac{{3+\sqrt{{29}}}}{2}&\text{ or }a=\displaystyle \frac{{3-\sqrt{{29}}}}{2}\end{align*}

$\scriptsize a=\displaystyle \frac{{3+\sqrt{{29}}}}{2}$:
\scriptsize \begin{align*}\text{LHS}&=3{{a}^{2}}-9a\\&=3{{\left( {\displaystyle \frac{{3+\sqrt{{29}}}}{2}} \right)}^{2}}-9\left( {\displaystyle \frac{{3+\sqrt{{29}}}}{2}} \right)\\&=3\left( {\displaystyle \frac{{9+6\sqrt{{29}}+29}}{4}} \right)-9\left( {\displaystyle \frac{{3+\sqrt{{29}}}}{2}} \right)\\&=\displaystyle \frac{{27+18\sqrt{{29}}+87}}{4}-\displaystyle \frac{{54+18\sqrt{{29}}}}{4}\\&=\displaystyle \frac{{27+18\sqrt{{29}}+87-54-18\sqrt{{29}}}}{4}\\&=\displaystyle \frac{{60}}{4}\\&=15\\&=\text{RHS}\end{align*}

$\scriptsize x=\displaystyle \frac{{3-\sqrt{{29}}}}{2}$:
\scriptsize \begin{align*}\text{LHS}&=3{{a}^{2}}-9a\\&=3{{\left( {\displaystyle \frac{{3-\sqrt{{29}}}}{2}} \right)}^{2}}-9\left( {\displaystyle \frac{{3-\sqrt{{29}}}}{2}} \right)\\&=3\left( {\displaystyle \frac{{9-6\sqrt{{29}}+29}}{4}} \right)-9\left( {\displaystyle \frac{{3-\sqrt{{29}}}}{2}} \right)\\&=\displaystyle \frac{{27-18\sqrt{{29}}+87}}{4}-\displaystyle \frac{{54-18\sqrt{{29}}}}{4}\\&=\displaystyle \frac{{27-18\sqrt{{29}}+87-54=18\sqrt{{29}}}}{4}\\&=\displaystyle \frac{{60}}{4}\\&=15\\&=\text{RHS}\end{align*}

Both solutions are valid. Therefore, $\scriptsize a=\displaystyle \frac{{3\pm \sqrt{{29}}}}{2}$.

### Exercise 2.1

1. Solve for the unknown in each case by completing the square:
1. $\scriptsize {{a}^{2}}+3a=-2$
2. $\scriptsize {{x}^{2}}-10x+2=0$
3. $\scriptsize {{y}^{2}}=7-8y$
4. $\scriptsize 2{{x}^{2}}=13x$
2. Solve the equation in 1.d. WITHOUT completing the square.

The full solutions are at the end of the unit.

## Summary

In this unit you have learnt the following:

• How to solve quadratic equations by completing the square.

# Unit 2: Assessment

#### Suggested time to complete: 30 minutes

1. Solve for the unknown in each of the following by completing the square:
1. $\scriptsize {{x}^{2}}+12x-3=0$
2. $\scriptsize 3{{a}^{2}}+a-4=0$
3. $\scriptsize 3(6x+{{x}^{2}})=-5$
4. $\scriptsize \displaystyle \frac{{4x}}{{(x+3)}}-\displaystyle \frac{{2x}}{{(x-4)}}=\displaystyle \frac{7}{{{{x}^{2}}-x-12}}$
2. Solve for $\scriptsize x$ in terms of $\scriptsize a$: $\scriptsize {{x}^{2}}-6x=2a$

The full solutions are at the end of the unit.

# Unit 2: Solutions

### Exercise 2.1

1. .
1. .
\scriptsize \begin{align*} a^2+3a&=-2\\ \therefore a^2+3a+2&=0\\ \therefore a^2+3a+\displaystyle \frac{9}{4}-\displaystyle \frac{9}{4}+2&=0\\ \therefore \left(a+\displaystyle \frac{3}{2}\right)^2&=\displaystyle \frac{1}{4}\\ \therefore a+\displaystyle \frac{3}{2}&=\pm\displaystyle \frac{1}{2}\\ \therefore a=-\displaystyle \frac{3}{2}+\displaystyle \frac{1}{2}&\text{ or }a=-\displaystyle \frac{3}{2}-\displaystyle \frac{1}{2}\\ \therefore a=-1&\text{ or }a=-2 \end{align*}
2. .
\scriptsize \begin{align*}{{x}^{2}}-10x+2 & =0\\\therefore {{x}^{2}}-10x+25-25+2 & =0\\\therefore {{(x-5)}^{2}}-23 & =0\\\therefore {{(x-5)}^{2}} & =23\\\therefore x & =5\pm \sqrt{{23}}\end{align*}
3. .
\scriptsize \begin{align*} y^2&=7-8y\\ \therefore y^2+8y-7&=0\\ \therefore y^2+8y+16-16-7&=0\\ \therefore (y+4)^2-23&=0 \therefore (y+4)^2&=23\\ \therefore y&=-4\pm\sqrt{23} \end{align*}
4. .
\scriptsize \begin{align*} 2x^2&=13x\\ \therefore 2x^2-13x&=0\\ \therefore 2\left(x^2-\displaystyle \frac{13}{2}x\right)&=0\\ \therefore x^2-\displaystyle \frac{13}{2}x&=0\\ \therefore x^2-\displaystyle \frac{13}{2}+\displaystyle \frac{169}{4}-\displaystyle \frac{169}{4}&=0\\ \therefore \left(x-\displaystyle \frac{13}{4}\right)^2&=\displaystyle \frac{169}{4}\\ \therefore x-\displaystyle \frac{13}{4}&=\pm\displaystyle \frac{13}{4}\\ \therefore x&=\displaystyle \frac{13}{4}\pm\displaystyle \frac{13}{4}\\ \therefore x=0&\text{ or }x=\displaystyle \frac{13}{2} \end{align*}
2. .
\scriptsize \begin{align*}2{{x}^{2}} & =13x\\\therefore 2{{x}^{2}}-13x & =0\\\therefore x(2x-13) & =0\\\therefore x=0\text{ } & \text{or }x=\displaystyle \frac{{13}}{2}\end{align*}

Back to Exercise 2.1

### Unit 2: Assessment

1. .
1. .
\scriptsize \begin{align*} x^2+12x-3&=0\\ \therefore x^2+12x+36-36-3&=0\\ \therefore (x+6)^2-39&=0\\ \therefore (x+6)^2&=39\\ \therefore x+6&=\pm\sqrt{39}\\ \therefore x&=-6\pm\sqrt{39} \end{align*}
2. .
\scriptsize \begin{align*} 3a^2+a-4&=0\\ \therefore 3\left(a^2+\displaystyle \frac{1}{3}a\right)&=4\\ \therefore a^2+\displaystyle \frac{1}{3}a&=\displaystyle \frac{4}{3}\\ \therefore a^2+\displaystyle \frac{1}{3}a+\displaystyle \frac{1}{36}&=\displaystyle \frac{4}{3}+\displaystyle \frac{1}{36}\\ \therefore \left(a+\displaystyle \frac{1}{6}\right)^2&=\displaystyle \frac{49}{36}\\ \therefore a+\displaystyle \frac{1}{6}&=\pm\displaystyle \frac{7}{6}\\ \therefore a=-\displaystyle \frac{1}{6}+\displaystyle \frac{7}{6}&\text{ or }a=-\displaystyle \frac{1}{6}-\displaystyle \frac{7}{6}\\ \therefore a=1&\text{ or }a=-\displaystyle \frac{8}{6}=-\displaystyle \frac{4}{3} \end{align*}
3. .
\scriptsize \begin{align*} 3(6x+x^2)&=-5\\ \therefore x^2+6x&=-\displaystyle \frac{5}{3}\\ \therefore x^2+6x+9&=-\displaystyle \frac{5}{3}+9\\ \therefore (x+3)^2&=\displaystyle \frac{22}{3}\\ \therefore x+3&=\pm\sqrt{\displaystyle \frac{22}{3}}\\ \therefore x&=-3\pm\sqrt{\displaystyle \frac{22}{3}} \end{align*}
4. .
\scriptsize \begin{align*}\displaystyle \frac{{4x}}{{(x+3)}}-\displaystyle \frac{{2x}}{{(x-4)}} & =\displaystyle \frac{7}{{{{x}^{2}}-x-12}}\\\therefore \displaystyle \frac{{4x}}{{(x+3)}}-\displaystyle \frac{{2x}}{{(x-4)}} & =\displaystyle \frac{7}{{(x+3)(x-4)}},x\ne -3,x\ne 4\\\therefore 4x(x-4)-2x(x+3) & =7\\\therefore 4{{x}^{2}}-16x-2{{x}^{2}}-6x-7 & =0\\\therefore 2{{x}^{2}}-22x-7 & =0\\\therefore 2({{x}^{2}}-11x) & =7\\\therefore {{x}^{2}}-11x & =\displaystyle \frac{7}{2}\\\therefore {{x}^{2}}-11x+\displaystyle \frac{{121}}{4} & =\displaystyle \frac{7}{2}+\displaystyle \frac{{121}}{4}\\\therefore {{\left( {x-\displaystyle \frac{{11}}{2}} \right)}^{2}} & =\displaystyle \frac{{135}}{4}\\\therefore x-\displaystyle \frac{{11}}{2} & =\displaystyle \frac{{\pm \sqrt{{135}}}}{2}\\\therefore x & =\displaystyle \frac{{11\pm \sqrt{{135}}}}{2}\end{align*}
2. .
\scriptsize \begin{align*} x^2-6x&=2a\\ \therefore x^2-6x+9&=2a+9\\ \therefore (x-3)^2&=2a+9\\ \therefore x-3&=\pm\sqrt{2a+9}\\ \therefore x&=3\pm\sqrt{2a+9} \end{align*}

Back to Unit 2: Assessment