Question

How do you solve `x^2+6x-4=0` by completing the square?

Answer 1

Force a perfect square trinomial on the left side of the equation. Take the square root of both sides and solve for `x`.

Explanation:

`x^2+6x-4=0`

Force a perfect square trinomial such that `a^2+2ab+b^2=(a+b)^2` on the left side of the equation. Then take the square root of both sides and solve for `x`.

Add `4` to both sides of the equation.

`x^2+6x=4`

Take the coefficient of the `x` term and divide it by `2` and square the result.

`6/2=3;` `3^2=9`

Add `9` to both sides.

`x^2+6x+9=4+9` =

`x^2+6x+9=13`

`a=x;` `b=3`

`(x+3)^2=13`

Take the square root of both sides.

`x+3=+-sqrt13`

Subtract `3` from both sides.

`x=-3+sqrt13`

`x=-3-sqrt13`