How do you solve #sqrt( 3x-3) + sqrt(2x+8) +1=0#?

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Answer 1 · 2016-04-06T00:30:00

Isolate one of the radicals on one side of the equation.

#sqrt(3x - 3) = -1 - sqrt(2x + 8)#

#(sqrt(3x - 3))^2 = (-1 - sqrt(2x + 8))^2#

#3x - 3 = 1 + 2x + 8 + 2sqrt(2x + 8)#

#x - 12 = 2sqrt(2x + 8)#

#(x - 12)^2 = (2sqrt(2x + 8))^2#

#x^2 - 24x + 144 = 4(2x + 8)#

#x^2 - 24x + 144 - 8x - 32 = 0#

#x^2 - 32x + 112 = 0#

#(x - 28)(x - 4) = 0#

#x = 28 and 4#

Checking both these solutions in the original equation we find that neither work. Therefore, there is no solution #(O/)#

Hopefully this helps!