Question

How do you solve `x² - 2x = 15` by completing the square?

Answer 1

`x=5` or `x=-3`

Explanation:

Although there is a better method, completing the square method is as such:

rearrange the formula to equate 0:

`x^{2}-2x-15=0`

`(x-1)^2-1-15=0`

Rearrange to get:

`(x-1)^{2}=16`

Square root on both sides to get:
`(x-1) = +-4`

Therefore adding 1 to both sides gives us either: `x=5` or `x=-3`

Answer 2

`x=5` and `x=-3`

Explanation:

`x^2-2x-15=0` is in the format of `ax^2+bx+c=0`.

As `c` has already been moved to the left, let's add `(b/2)^2` to both sides and factor:

`x^2-2x=15`
`x^2-2x+(b/2)^2=15+(b/2)^2`

Your value of `b` is the coefficient before the `x` in `bx` (from `ax^2+bx+c=0`):

`x^2-2x+(-2/2)^2=15+(-2/2)^2`
`x^2-2x+(-1)^2=15+(-1)^2`
`x^2-2x+1=15+1`
`(x-1)(x-1)=16`
`(x-1)^2=16`

Now solve for `x`:

`x-1=+-sqrt16`
`x-1=+-4`
`x=4+1` and `x=-4+1`

Therefore, `x=5` and `x=-3`

Answer 3

`x=-3, x=5`

Explanation:

`x^2-2x-15=0`

`rArr (x-1)^2-1-15`

`rArr (x-1)^2-16=0`

`rArr (x-1)^2=16`

`rArr x-1=pmsqrt16`

`rArr x-1=pm4`

`rArr x=1pm4`

`rArr x=1-4` or `x=1+4`

`rArr x=-3, x=5`