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

Question

6616 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-16T18:06:53

$x=3+sqrt13, 3-sqrt13$

$x^x-6x-4=0$

Completing the square means that we will force a perfect square trinomial on the left side of the equation, then solve for $x$ .

Add $4$ to both sides of the equation.

$x^2-6x=4$

Divide the coefficient of the $x$ term by $2$ , then square the result. Add it 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 in the form $a^2-2ab+b^2=(a-b)^2$ , where $a=x and b=3$ .

Substitute $(x-3)^2$ for $x^2-6x+9$ .

$(x-3)^2=13$

Take the square root of both sides.

$x-3=+-sqrt13$

Add $3$ to both sides of the equation.

$x=3+-sqrt13$

Solve for $x$ .

$x=3+sqrt13$

$x=3-sqrt13$

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