How do you solve # x^2+6x-1=0#?

1 Answer
Jun 14, 2015

Answer:

Solve this equation by completing the square.

#x=sqrt10-3#

#x=-sqrt10-3#

Explanation:

Solve #x^2+6x-1=0#

We are going to create a perfect square trinomial on the left side of the equation. A perfect square trinomial has the form #a^2+2ab+b^2=(a+b)^2#. Then we can factor the perfect square trinomial and solve for #x#.

Add the constant term #(1)# to both sides of the equation.

#x^2+6x=1#

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

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

#x^2+6x+9=10#

We now have a perfect square trinomial on the left side which can be factored to #(x+3)^2#.

#(x+3)^2=10#

Take the square root of both sides.

#x+3=+-sqrt 10#

#x=+-sqrt10-3#

#x=sqrt10-3#

#x=-sqrt10-3#