Mathematics » Equations and Inequalities » Completing The Square

# Completing the Square

## Optional Investigation

### Completing the square

Can you solve each equation using two different methods?

1. $$x^2 – 4 = 0$$
2. $$x^2 – 8 = 0$$
3. $$x^2 -4x + 4 = 0$$
4. $$x^2 -4x – 4 = 0$$

Factorising the last equation is quite difficult. Use the previous examples as a hint and try to create a difference of two squares.

We have seen that expressions of the form $$x^2 – b^2$$ are known as differences of squares and can be factorised as $$(x-b)(x+b)$$. This simple factorisation leads to another technique for solving quadratic equations known as completing the square.

Consider the equation $$x^2-2x-1=0$$. We cannot easily factorise this expression. When we expand the perfect square $$(x-1)^2$$ and examine the terms we see that $$(x-1)^2 = x^2-2x+1$$.

We compare the two equations and notice that only the constant terms are different. We can create a perfect square by adding and subtracting the same amount to the original equation.

\begin{align*} x^2-2x-1 &= 0 \\ (x^2-2x+1)-1-1 &= 0 \\ (x^2-2x+1)-2 &= 0 \\ (x-1)^2-2 &= 0 \end{align*}

Method 1: Take square roots on both sides of the equation to solve for $$x$$. \begin{align*} (x-1)^2-2 &= 0 \\ (x-1)^2 &= 2 \\ \sqrt{(x-1)^2} &= \pm \sqrt{2} \\ x-1 &= \pm \sqrt{2} \\ x &= 1 \pm \sqrt{2} \\ \text{Therefore }x &= 1 + \sqrt{2} \text{ or }x = 1 – \sqrt{2} \end{align*}

Very important: Always remember to include both a positive and a negative answer when taking the square root, since $$2^2 = 4$$ and $$(-2)^2 = 4$$.

Method 2: Factorise the expression as a difference of two squares using $$2 = (\sqrt{2})^2$$.

We can write

\begin{align*} (x-1)^2-2 &= 0 \\ (x-1)^2 – ( \sqrt{2} )^2 &= 0 \\ ( (x-1) + \sqrt{2} )( (x-1) – \sqrt{2} ) &= 0 \end{align*}

The solution is then \begin{align*} (x-1) + \sqrt{2} &= 0 \\ x &= 1 – \sqrt{2} \end{align*} or \begin{align*} (x-1) – \sqrt{2} &= 0 \\ x &= 1 + \sqrt{2} \end{align*}

Method for solving quadratic equations by completing the square

1. Write the equation in the standard form $$a{x}^{2}+bx+c=0$$.

2. Make the coefficient of the $${x}^{2}$$ term equal to $$\text{1}$$ by dividing the entire equation by $$a$$.

3. Take half the coefficient of the $$x$$ term and square it; then add and subtract it from the equation so that the equation remains mathematically correct. In the example above, we added $$\text{1}$$ to complete the square and then subtracted $$\text{1}$$ so that the equation remained true.

4. Write the left hand side as a difference of two squares.

5. Factorise the equation in terms of a difference of squares and solve for $$x$$.

## Example

### Question

Solve by completing the square: $$x^2-10x-11=0$$

### Make sure the coefficient of the $$x^2$$ term is equal to $$\text{1}$$

$x^2-10x-11=0$

### Take half the coefficient of the $$x$$ term and square it; then add and subtract it from the equation

The coefficient of the $$x$$ term is $$-\text{10}$$. Half of the coefficient of the $$x$$ term is $$-\text{5}$$ and the square of it is $$\text{25}$$. Therefore $$x^2 – 10x + 25 – 25 – 11 = 0$$.

### Write the trinomial as a perfect square

\begin{align*} (x^2 – 10x + 25) – 25 – 11 &= 0 \\ (x-5)^2 – 36 &= 0 \end{align*}

### Method 1: Take square roots on both sides of the equation

\begin{align*} (x-5)^2 – 36 &= 0 \\ (x-5)^2 &= 36 \\ x-5 &= \pm\sqrt{36} \end{align*}

Important: When taking a square root always remember that there is a positive and negative answer, since $$(6)^2 = 36$$ and $$(-6)^2 = 36$$.

$x – 5 = \pm 6$

### Solve for $$x$$

$x = -1 \text{ or } x = 11$

### Method 2: Factorise equation as a difference of two squares

\begin{align*} (x-5)^2 – (6)^2 &= 0 \\ [(x-5) + 6] [(x-5) – 6] &= 0 \end{align*}

### Simplify and solve for $$x$$

\begin{align*} (x+1)(x-11) &= 0 \\ \therefore x = -1 \text{ or } x &= 11 \end{align*}

$x = -1 \text{ or } x = 11$

Notice that both methods produce the same answer. These roots are rational because $$\text{36}$$ is a perfect square.

## Example

### Question

Solve by completing the square: $$2x^2 – 6x – 10 = 0$$

### Make sure that the coefficient of the $$x^2$$ term is equal to $$\text{1}$$

The coefficient of the $${x}^{2}$$ term is $$\text{2}$$. Therefore divide the entire equation by $$\text{2}$$:

$x^2 – 3x – 5 = 0$

### Take half the coefficient of the $$x$$ term, square it; then add and subtract it from the equation

The coefficient of the $$x$$ term is $$-\text{3}$$, so then $$( \dfrac{-3}{2} )^2 = \dfrac{9}{4}$$:

$( x^2 – 3x + \cfrac{9}{4}) – \cfrac{9}{4} – 5 = 0$

### Write the trinomial as a perfect square

\begin{align*} ( x – \cfrac{3}{2} )^2 – \cfrac{9}{4} – \cfrac{20}{4} &= 0 \\ ( x – \cfrac{3}{2} )^2 – \cfrac{29}{4} &= 0 \end{align*}

### Method 1: Take square roots on both sides of the equation

\begin{align*} ( x – \cfrac{3}{2} )^2 – \cfrac{29}{4} &= 0 \\ ( x – \cfrac{3}{2} )^2 &= \cfrac{29}{4} \\ x – \cfrac{3}{2} &= \pm \sqrt{\cfrac{29}{4}} \end{align*}

Remember: When taking a square root there is a positive and a negative answer.

### Solve for $$x$$

\begin{align*} x – \cfrac{3}{2} &= \pm \sqrt{\cfrac{29}{4}} \\ x &= \cfrac{3}{2} \pm \cfrac{\sqrt{29}}{2} \\ &= \cfrac{3 \pm \sqrt{29}}{2} \end{align*}

### Method 2: Factorise equation as a difference of two squares

\begin{align*} ( x – \cfrac{3}{2} )^2 – \cfrac{29}{4} &= 0 \\ ( x – \cfrac{3}{2} )^2 – ( \sqrt{\cfrac{29}{4}} )^2 &= 0\\ ( x – \cfrac{3}{2} – \sqrt{\cfrac{29}{4}} ) ( x – \cfrac{3}{2} + \sqrt{\cfrac{29}{4}} ) &= 0 \end{align*}

### Solve for $$x$$

\begin{align*} ( x – \cfrac{3}{2} – \cfrac{\sqrt{29}}{2} ) ( x – \cfrac{3}{2} + \cfrac{\sqrt{29}}{2} ) &= 0 \\ \text{Therefore } x = \cfrac{3}{2} + \cfrac{\sqrt{29}}{2} &\text{ or } x = \cfrac{3}{2} – \cfrac{\sqrt{29}}{2} \end{align*}

Notice that these roots are irrational since $$\text{29}$$ is not a perfect square.