How do you solve #-3( x + 6) ^ { 2} + 12= 0#?

Question

505 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-01T19:24:05

#x = -4" " or" " x =-8#

The equation is somewhat easier than it first appears because there is no separate #x# term in the original equation.

Just as #x^2 = 25# can be solved as #x = +-sqrt25#,

we can use the same approach in this case.

#-3(x+6)^2 +12 =0#

#-3(x+6)^2 = -12" "larr div -3#

#(-3(x+6)^2)/-3 = (-12)/-3#

#(x+6)^2 = 4" "larr# find the square root of both sides:

#x +6 = +-sqrt4#

#x = +2-6 = -4#

#x= -2-6 = -8#