How do you solve #3x^2+12x+81=15# by completing the square?

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Answer 1 · 2017-06-30T09:25:45

#x = -2+-i sqrt(22)#

Note that all of the terms are divisible by #3#, so let's divide both sides of the equation by #3# to get:

#x^2+4x+27=5#

To make the left hand side into a perfect square, subtract #23# from both sides to get:

#x^2+4x+4 = -22#

The left hand side is now #(x+2)^2#, so our equation is:

#(x+2)^2 = -22#

The square of any real number is non-negative, so this quadratic equation only has non-real complex solutions.

If you want to proceed further, note that if #n > 0# then:

#sqrt(-n) = i sqrt(n)#

where #i# is the imaginary unit, satisfying #i^2 = -1#.

So we find:

#(x+2)^2 = (i sqrt(22))^2#

and hence:

#x+2 = +-i sqrt(22)#

Subtracting #2# from both sides we get:

#x = -2+-i sqrt(22)#