Question

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

Answer 1

`x=3+sqrt13, 3-sqrt13`

Explanation:

`6x+4=x^2`

Completing the square means forcing a perfect square trinomial on the left side of the equation.

Bring all terms to the left side.

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

Multiply boths sides times `-1`.

`-1(-x^2+6x+4=0)` =

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

Add `4` to both sides of the equation.

`x^2-6x=4`

Divide the x-term by 2 and square the result. Add to both sides of the equation.

`((-6)/2)^2=-3^2=9`

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

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

We now have a perfect square trinomial on the left side. `a^2-2ab+b^2=(a-b)^2`, where `a=x and b=3`.

Substitute `(x-3)^2` into the equation.

`(x-3)^2=13` =

Take the square root of both sides and solve for `x`.

`x-3=+-sqrt13`

Add `3` to both sides.

`x=3+-sqrt13`

Solve for `x`.

`x=3+sqrt13`

`x=3-sqrt13`