How do you solve #sqrt(x-4) – sqrt(x+8)=2#?

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Answer 1 · 2016-02-08T02:29:00

Square both sides of the equation, and isolate one radical to one side of the equation (see solution below).

#sqrt(x - 4) = 2 + sqrt(x + 8)#

#(sqrt(x - 4))^2 = (2 + sqrt(x + 8))^2#

#x - 4 = 4 + 4sqrt(x + 8) + x + 8#

#-4 - 4 - 8 = 4sqrt(x + 8)#

#-16 = 4sqrt(x + 8)#

#-4 = sqrt(x + 8)#

Square both sides again to get rid of the remaining square root.

#(-4)^2 = (sqrt(x + 8))^2#

16 = x + 8

8 = x

Check in the original equation to make sure the solution is not an extraneous root.

#sqrt(8 - 4) - sqrt(8 + 8) != 2#

So, this equation has no solution (#O/#)

Practice exercises:

a) #sqrt(2x + 6) - sqrt(x + 1) = 2#

b) #sqrt(2x - 1) + sqrt(9x + 4) = 10#

Good luck!