Question

How do you solve `x^2 + 6x - 9 = 0` by completing the square?

Answer 1

`x = -3 ± 3sqrt2`

Explanation:

For a polynomial `x^2+bx+c`, the completed square is of the form `(x+b/2)^2-(b/2)^2+c`

`therefore x^2+6x-9 = (x+3)^2-9-9 = (x+3)^2-18`

`(x+3)^2-18 = 0`

`(x+3)^2=18`

`x+3 = ±sqrt18 = ±3sqrt2`

`x = -3 ± 3sqrt2`