How do you solve #x^2+10x-2=0# by completing the square?

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Answer 1 · 2015-07-08T07:36:13

Force a perfect square trinomial on the left side. Take the square root of both sides and solve for#x#.

#x^2+10x-2=0#

Add #2# to both sides of the equation.

#x^2+10x=2#

Force a perfect square trinomial on the left side of the equation.

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

#(10/2)^2=5^2=25#

#x^2+10x+25=2+25# =

#x^2+10x+25=27#

We now have a perfect square trinomial on the left side with the form #a^2+2ab+b^2=(a+b)^2#.

#a=x,# #b=5#

#(x+5)^2=27#

Take the square root of both sides.

#(x+5)=+-sqrt 27# =

#x+5=+-sqrt 3sqrt 9# =

#x+5=+-3sqrt 3# =

Subtract #5# from both sides.

#x=-5+-3sqrt3#

#x=-5+3sqrt 3#

#x=-5-3sqrt3#

#x=-5+3sqrt 3, -5-3sqrt3#