How do you solve by completing the square #x^2+2x-5=0#?

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Answer 1 · 2015-06-30T20:02:37

Force a perfect square trinomial on the left side. Take the square root of both sides. Solve for #x#, which will have two values.

Completing the square involves forcing a perfect square trinomial on the left side of the equation, then solving for #x#.

The form for a perfect square is #a^2+2ab+b^2=(a+b)^2#.

#x^2+2x-5=0#.

Add #5# to both sides of the equation.

#x^2+2x=5#

Divide the coefficient of the #x# value by #2#, then square the result.

#2/2=1;# #1^2=1#

Add the result to both sides.

#x^2+2x+1=5+1# =

#x^2+2x+1=6#

The left side is now a perfect square trinomial.

#x^2+2x+1=(x+1)^2#

#(x+1)^2=6#

Take the square root on both sides.

#x+1=+-sqrt6#

Subtract #1# from both sides.

#x=sqrt6-1#

#x=-sqrt6-1#