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# 2.2 Completing the square

## 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 $$2 = \left(\sqrt{2}\right)^2$$.

We can write

\begin{align*} (x-1)^2-2 &= 0 \\ (x-1)^2 - \left( \sqrt{2} \right)^2 &= 0 \\ \left( (x-1) + \sqrt{2} \right)\left( (x-1) - \sqrt{2} \right) &= 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 $$x$$.

## Worked example 6: Solving quadratic equations by completing the square

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 \\ \left[\left(x-5\right) + 6\right] \left[\left(x-5\right) - 6\right] &= 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.

## Worked example 7: Solving quadratic equations by completing the square

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 $$\left( \dfrac{-3}{2} \right)^2 = \dfrac{9}{4}$$:

$\left( x^2 - 3x + \frac{9}{4}\right) - \frac{9}{4} - 5 = 0$

### Write the trinomial as a perfect square

\begin{align*} \left( x - \frac{3}{2} \right)^2 - \frac{9}{4} - \frac{20}{4} &= 0 \\ \left( x - \frac{3}{2} \right)^2 - \frac{29}{4} &= 0 \end{align*}

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

\begin{align*} \left( x - \frac{3}{2} \right)^2 - \frac{29}{4} &= 0 \\ \left( x - \frac{3}{2} \right)^2 &= \frac{29}{4} \\ x - \frac{3}{2} &= \pm \sqrt{\frac{29}{4}} \end{align*}

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

### Solve for $$x$$

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

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

\begin{align*} \left( x - \frac{3}{2} \right)^2 - \frac{29}{4} &= 0 \\ \left( x - \frac{3}{2} \right)^2 - \left( \sqrt{\frac{29}{4}} \right)^2 &= 0\\ \left( x - \frac{3}{2} - \sqrt{\frac{29}{4}} \right) \left( x - \frac{3}{2} + \sqrt{\frac{29}{4}} \right) &= 0 \end{align*}

### Solve for $$x$$

\begin{align*} \left( x - \frac{3}{2} - \frac{\sqrt{29}}{2} \right) \left( x - \frac{3}{2} + \frac{\sqrt{29}}{2} \right) &= 0 \\ \text{Therefore } x = \frac{3}{2} + \frac{\sqrt{29}}{2} &\text{ or } x = \frac{3}{2} - \frac{\sqrt{29}}{2} \end{align*}

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

## Solution by completing the square

Textbook Exercise 2.2

Solve the following equations by completing the square:

$${x}^{2}+10x-2=0$$

\begin{align*} x^2 + 10x - 2 &= 0\\ x^2 +10x&= 2\\ x^2 +10x +25 &= 2 + 25\\ (x+5)^2 - 27 &= 0\\ \left[(x+5) + \sqrt{27}\right]\left[(x+5) - \sqrt{27}\right] &= 0\\ (x+5) = -\sqrt{27} &\text{ or } (x+5) = \sqrt{27}\\ x = -5 - 3\sqrt{3} &\text{ or } x= -5 + 3\sqrt{3} \end{align*}

$${x}^{2}+4x+3=0$$

\begin{align*} x^2+4x+3&=0\\ x^2 +4x &= -3\\ x^2 + 4x + 4 &= -3 + 4\\ (x+2)^2 &= 1\\ (x+2)&= \pm \sqrt {1}\\ &=\pm 1\\ x = -2 + 1 = -1 &\text{ or } x = -2-1=-3 \end{align*}

$$p^2 - 5 = - 8p$$

\begin{align*} p{^2} - 5 &= - 8p\\ p{^2}+8p-5 &= 0\\ p^2 + 8p &=5\\ p^2 + 8p + 16 &= 5 + 16\\ (p+4)^2 &= 21\\ (p+4)&=\pm \sqrt{21}\\ p &= -4 \pm \sqrt{21} \end{align*}

$$2(6x + x^2) = -4$$

\begin{align*} 2(6x + x^2) &= -4 \\ 2x^2+12x+4 &= 0\\ x^2 + 6x + 2 &= 0\\ x^2 +6x &= -2\\ x^2 + 6x + 9&= -2 +9\\ (x+3)^2 &= 7\\ x + 3 &= \pm \sqrt{7}\\ x &= -3 \pm \sqrt{7} \end{align*}

$${x}^{2}+5x+9=0$$

\begin{align*} x^2+5x+9 &= 0\\ x^2 + 5x &= -9\\ x^2 + 5x + \frac{25}{4} &= -9 + \frac{25}{4}\\ \left(x+\frac{5}{2}\right)^2 &= -\frac{11}{4}\\ x + \frac{5}{2} &= \pm \sqrt{-\frac{11}{4}}\\ \text{ No real solution } \end{align*}

$$t^2 + 30 = 2(10-8t)$$

\begin{align*} t^2 + 30 &= 2(10 - 8t) \\ t^2+16t+10&=0\\ t^2 +16t &= -10\\ t^2 + 16t+64 &= -10 +64\\ (t+8)^2&=54\\ t + 8 &= \pm \sqrt{54}\\ t &= -8 \pm \sqrt{9 \times 6} \\ \therefore t &= -8 \pm 3\sqrt{6} \end{align*}

$$3{x}^{2}+6x-2=0$$

\begin{align*} 3x^2+6x-2&=0\\ x^2 + 2x &= \frac{2}{3}\\ x^2 + 2x + 1 &= \frac{2}{3} + 1\\ (x+1)^2 &= \frac{5}{3}\\ x +1 &= \pm \sqrt{\frac{5}{3}}\\ x &= -1 \pm \sqrt{\frac{5}{3}} \end{align*}

$${z}^{2}+8z-6=0$$

\begin{align*} z^2 + 8z - 6 &= 0\\ z^2 +8z &= 6\\ z^2 + 8z+16 &= 6+16\\ (z+4)^2 &= 22\\ z+4 &= \pm \sqrt{22}\\ z &= -4 \pm \sqrt{22} \end{align*}

$$2z^2 = 11z$$

\begin{align*} 2z^2 &= 11z\\ 2z^2 - 11z &= 0\\ z^2 - \frac{11}{2}z &= 0\\ z^2 - \frac{11}{2}z + \frac{121}{16} &= \frac{121}{16}\\ \left(z-\frac{11}{4}\right)^2 &= \frac{121}{16}\\ z - \frac{11}{4} &= \pm \frac{11}{4}\\ z = \frac{11}{4} + \frac{11}{4} = \frac{11}{2} &\text{ or } z = \frac{11}{4} - \frac{11}{4} = 0 \end{align*}

$$5+4z-{z}^{2}=0$$

\begin{align*} 5 + 4z - z^2 &= 0\\ z^2 -4z &= 5\\ z^2 - 4z +4 &= 5+4\\ (z-2)^2 &= 9\\ z-2 &= \pm \sqrt{9}\\ z = 2+3 = 5 &\text{ or } z = 2-3 = -1 \end{align*}

Solve for $$k$$ in terms of $$a$$: $$k^2 + 6k+ a = 0$$

\begin{align*} k^2 + 6k+ a &= 0\\ k^2 + 6k &= -a\\ k^2 + 6k + 9 &= 9 - a\\ (k + 3)^2 &= 9 - a\\ k + 3 &= \pm \sqrt{9 - a}\\ k & = -3 \pm \sqrt{9 - a} \end{align*}

Solve for $$y$$ in terms of $$p$$, $$q$$ and $$r$$: $$py^2 + qy + r = 0$$

\begin{align*} py^2 + qy + r &= 0\\ y^2 + \frac{q}{p}y + \frac{r}{p} &= 0\\ y^2 + \frac{q}{p}y &= -\frac{r}{p}\\ y^{2} + \frac{q}{p}y + \left(\frac{q}{2p}\right)^{2} &= \left(\frac{q}{2p}\right)^{2} - \frac{r}{p}\\ \left(y + \frac{q}{2p}\right)^2 &= \frac{q^{2}}{4p^{2}} - \frac{r}{p} \\ y + \frac{q}{2p} &= \pm \sqrt{\frac{q^{2} - 4pr}{4p^{2}}}\\ y &= -\frac{q}{2p} \pm \frac{\sqrt{q^{2} - 4pr}}{2p}\\ y & = \frac{-q \pm \sqrt{q^{2} - 4pr}}{2p} \end{align*}