How do you solve #x+3=sqrt(18-2x^2)# algebraically?

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Answer 1 · 2016-12-26T19:26:05

#{-3}#

Square both sides.

#(x + 3)^2 = (sqrt(18 - 2x^2))^2#

#x^2 + 6x + 9 = 18 - 2x^2#

#3x^2 + 6x - 9 = 0#

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

#x^2 + 2x - 3 = 0#

#(x + 3)(x -1) = 0#

#x = -3 and 1#

Checking in the original equation, you will find that only #x = -3# will work.

Hopefully this helps!