How do you solve # c^2+6c-27=0# by completing the square?

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Answer 1 · 2015-05-07T21:07:41

The answer is #c=3#, #c=-9# .

Solve #c^2+6c-27=0# by completing the square.

Add #27# to both sides.

#c^2+6c=27#

Halve the coefficient of #6c# then square it.
#6/2=3#

#3^2=9#

Add #9# to both sides of the equation.

#c^2+6c+9=27+9# =

#c^2+6c+9=36#

The lefthand side of the equation is now a perfect trinomial square. Factor the perfect trinomial square. #a^2+2ab+b^2=(a+b)^2#

#a=c#, #b=3#

#(c+3)^2=36#

Take the square root of both sides and solve for #c#.

#c+3=+-sqrt36#

#c+3=+-6#

When #c+3=6#:

#c=6-3#, #c=3#

When #c+3=-6#:

#c=-6-3#, #c=-9#

Check:

#3^2+6*3-27=0# =

#9+18-27=0# =

#27-27=0#

#-9^2+6(-9)-27=0# =

#81-54-27=0# =

#81-81=0#