How do you solve #x^2+4x-1=0# by completing the square?

1 Answer
Jul 10, 2015

Answer:

#x=-2+sqrt5,# #-2-sqrt5#

Explanation:

#x^2+4x-1=0#

Add #1# to both sides of the equation.

#x^2+4x=1#

Force a perfect square trinomial on the left side by dividing the coefficient of the #x# term and squaring the result. Add it to both sides of the equation.

#(4/2)^2=2^2=4#

#x^2+4x+4=1+4# =

#x^2+4x+4=5#

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

#(x+2)^2=5#

Take the square root of both sides.

#x+2=+-sqrt5#

Subtract #2# from both sides.

#x=-2+-sqrt5#

The two values for #x#.

#x=-2+sqrt5#

#x=-2-sqrt5#