WTMaths logo
Solving by Completing the Square

Solving by Completing the Square

Quadratic equations can be solved by completing the square. Get the equation in the form `ax^2^+bx+c` and then change the equation to completed square form.

Once the equation is in completed square form, solve the resulting equation. Note that when square rooting the equation, the square root will give both a positive and a negative value.

Once the answers have been obtained, check.

Example 1

By completing the square, solve `x^2 + 8x + 12 = 0`.

`x^2 +8x + 12 = 0`

Using the form `ax^2 + bx + c`

Half the value of the `b` term = 4

`(x+4)^2 = x^2 + 8x + 16`

Need to subtract 4 to make the original equation

`(x+4)^2 - 4 = x^2 +8x + 12 = 0`

Solve the equation `(x+4)^2 - 4` = 0
add 4 both sides `(x+4)^2` = 4
square root both sides `x+4` = ±2
subtract 4 both sides `x` = -2, -6

Check:

(-2)2 + 8(-2) + 12 = 0✔

(-6)2 + 8(-6) + 12 = 0✔

Answer: `x=-2` and `x=-6`

Example 2

Find the distinct solution for `9x^2 + 12x = 3` by completing the square.

Show your answer in surd form.

Rearrange the original expression to have zero on one side

`9x^2 + 12x - 3 = 0`

Find the square root of the `x^2` coefficient = 3

Divide the coefficient of the `x` term (12) by that three, then half = 2

The squared term is `(3x + 2)`

Multiply out `(3x + 2)^2` = `9x^2 + 12x + 4`

Adjust the expression by -7

`(3x + 2)^2 - 7 = 0`

Solve the equation `(3x+2)^2 - 7` `=` `0`
add 7 both sides `(3x+2)^2` `=` `7`
square root both sides `3x+2` `=` `±sqrt(7)`
subtract 2 both sides `3x` `=` `±sqrt(7)-2`
divide by 3 both sides `x` `=` `frac(±sqrt(7)-2)(3)`

Check:

`9(frac(sqrt(7)-2)(3))^2+12(frac(sqrt(7)-2)(3)) = 3`

`9(frac(-sqrt(7)-2)(3))^2+12(frac(-sqrt(7)-2)(3)) = 3`

Answer: `x=frac(±sqrt(7) - 2)(3)`

See also Completing the Square