## How to solve quadratics by completing the square

Completing the square is one of the most versatile technique in quadratic equations and expressions. By completing the square we can solve a quadratic function to find the:

- roots of the quadratic function
- turning points (maximum or minimum point) of the quadratic function

Before we start solving quadratic functions let us look at what a square is. A square is a resultant of multiplying two identical numbers. For example, 9 is a square obtained from multiplying 3 and 3.

Knowing the properties of a square number will help us understand how completing the square works. This tutorial is not about memorising the method by about understanding how the method actually works.

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## Squares of algebraic functions

Multiplying an algebraic function by itself will give us the square of the function. For example:

(x + 3)(x + 3) = (x + 3)^{2}

We can also expand the brackets by multiplying every term in the first bracket by every term in the second bracket to get:

(x + 3)(x + 3) = x^{2} + 3x + 3x + 9

- = x
^{2}+ 6x + 9

Now let us look at the relationship between the terms of the squared function and its corresponding expanded form.

(x + 3)^{2} = x^{2} + 6x + 9

- we have 3 in the squared function → (x +
**3**)^{2} - on the expanded function we have 2×3 as the coefficient of x → x
^{2}+**6**x + 9 - we have 3
^{2}as the last term → x^{2}+ 6x +**9**

This relationship applies to all quadratic perfect squares and let us prove it as follows:

(x + a)^{2} = (x + a)(x + a)

- = x
^{2}+ ax + ax + a^{2} - = x
^{2}+**2a**x +**a**^{2}

This is very important because it shows us that for a square (x^{2} + 2ax + a^{2}), the last term (a^{2}) is actually *the square of half the coefficient of x*. In short, a quadratic expression is a square if when you take the coefficient of x, divide it by 2 and square it you get the last term. Now lets look at the following examples and see if they are squares:

- x
^{2}+ 10x + 36- the coefficient of x is 10
- divide 10 by 2 to get 5
- square the 5 to get 25
- this means that the expression is not a perfect square and it would have been a perfect square if 36 had been replaced by 25.

- y
^{2}+ 8y + 16- the coefficient of y is 8
- divide 8 by 2 to get 4
- square the 4 to get 16
- therefore the expression is a perfect square.

## Completing the square

Completing the square is a process of converting a non square quadratic expression into a perfect square before we solve it.

### Example 1: Complete the square and factorise the following quadratic expressions.

- x
^{2}+ x - x
^{2}+ 2x - x
^{2}+ 3x - x
^{2}+ 4x - x
^{2}+ 5x

### Solution 1

If you have been following the tutorial very carefully you should now understand that we complete the square by *adding the square of half the coefficient of x*. This is what we are going to do and then we factorise the resultant quadratic function in the normal way.

x

^{2}+ x + 0.5^{2}- = x
^{2}+ x + 0.25 - = (x + 0.5)
^{2}

- = x
x

^{2}+ 2x + 1^{2}- = x
^{2}+ 2x + 1 - = (x + 1)
^{2}

- = x
x

^{2}+ 3x + 1.5^{2}- = x
^{2}+ 3x + 2.25 - = (x + 1.5)
^{2}

- = x
x

^{2}+ 4x + 2^{2}- = x
^{2}+ 4x + 4 - = (x + 2)
^{2}

- = x

After learning how to complete the square, let us now start solving the quadratic equations.

### Example 2: Solve the following quadratic equations by completing the square.

- x
^{2}+ 8x + 5 = 0 - 2x
^{2}+ 8x + 5 = 0

### Solution 2

$latex x^2 + 8x + 5 = 0$

$latex x^2 + 8x = -5$

$latex x^2 + 8x + 4^2= -5 + 4^2$

$latex x^2 + 8x + 4^2= 16-5$

$latex x^2 + 8x + 4^2= 11$

$latex (x + 4)^2= 11$

$latex \sqrt{(x + 4)^2}= \pm \sqrt{11}$

$latex x + 4= \pm \sqrt{11}$

$latex x = -4 \pm \sqrt{11}$

$latex x = -4 + \sqrt{11} \ or\ -4 – \sqrt{11}$

$latex x = -0.68 \ or\ -7.32 \ (to\ 2dp)$

$latex 2x^2 + 8x + 5 = 0$

$latex 2x^2 + 8x = -5$

$latex \frac{2x^2}{2} + \frac{8x}{2} = \frac{-5}{2}$

$latex x^2 + 4x = \frac{-5}{2}$

$latex x^2 + 4x + 2^2= \frac{-5}{2} + 2^2$

$latex (x + 2)^2 = 4 – \frac{5}{2}$

$latex (x + 2)^2 = \frac{8 – 5}{2}$

$latex (x + 2)^2 = \frac{3}{2}$

$latex \sqrt{(x + 2)^2} = \pm \sqrt{\frac{3}{2}}$

$latex x + 2 = \pm \sqrt{\frac{3}{2}}$

$latex x =-2 \pm \sqrt{\frac{3}{2}}$

$latex x = -3.22 or -0.78\ (\ to \ 2dp)$

Completing the square can also be used to derive the general quadratic formula.

### Example 3: Derive the quadratic formula by completing the square

$latex ax^2 + bx + c = 0$

$latex ax^2 + bx = -c$

$latex \frac{ax^2}{a} + \frac{bx}{a} = \frac{-c}{a}$

$latex x^2 + \frac{bx}{a} + (\frac{b}{2a})^2 = \frac{-c}{a} + (\frac{b}{2a})^2$

$latex (x + \frac{b}{2a})^2 = \frac{b^2}{4a^2} + \frac{-c}{a} $

$latex (x + \frac{b}{2a})^2 = \frac{b^2}{4a^2} – \frac{c}{a} $

$latex (x + \frac{b}{2a})^2 = \frac{b^2 – 4ac}{4a^2}$

$latex \sqrt{(x + \frac{b}{2a})^2} = \pm \sqrt{\frac{b^2 – 4ac}{4a^2}}$

$latex x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 – 4ac}}{2a}$

$latex x = – \frac{b}{2a} \pm \frac{\sqrt{b^2 – 4ac}}{2a}$

$latex x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$

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