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

Question

5346 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-11T14:32:58

$x = -3 ± 3sqrt2$

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

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

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

$(x+3)^2=18$

$x+3 = ±sqrt18 = ±3sqrt2$

$x = -3 ± 3sqrt2$

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