How do you solve the quadratic equation by completing the square: x^2-6x-8=0x26x8=0?

1 Answer
Jul 14, 2015

x=3+sqrt(17), 3-sqrt(17)x=3+17,317

Explanation:

x^2-6x-8=0x26x8=0

In order to solve a quadratic equation by completing the square, we must force a perfect square trinomial on the left side. a^2+2ab+b^2=(a+b)^2a2+2ab+b2=(a+b)2

Add 88 to both sides of the equation.

x^2-6x=8x26x=8

Divide the coefficient of the xx term by 22 and square the result. Add it to both sides of the equation.

((-6)/(2))^2=(-3)^2=9(62)2=(3)2=9

x^2-6x+9=8+9x26x+9=8+9 =

x^2-6x+9=17x26x+9=17

We now have a perfect square trinomial on the left side, in which
a=x,a=x, and b=-3b=3.

Now we can solve for xx.

(x-3)^2=17(x3)2=17

Take the square root of both sides.

(x-3)=+-sqrt(17)(x3)=±17

x=3+-sqrt(17)x=3±17

x=3+sqrt(17)x=3+17

x=3-sqrt(17)x=317