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

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Answer 1 · 2015-07-14T23:00:50

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

#x^2-6x-8=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)^2#

Add #8# to both sides of the equation.

#x^2-6x=8#

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

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

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

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

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

Now we can solve for #x#.

#(x-3)^2=17#

Take the square root of both sides.

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

#x=3+-sqrt(17)#

#x=3+sqrt(17)#

#x=3-sqrt(17)#