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

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Answer 1 · 2016-03-15T14:38:21

Rearrange and square a couple of times to find:

#x = 8#

Check the answer since squaring can introduce spurious solutions.

Add #sqrt(x-4)# to both sides to get:

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

Square both sides (noting that this may introduce spurious solutions):

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

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

Subtract #x# from both sides and transpose to get:

#4 sqrt(x-4) = 8#

Divide both sides by #4# to get:

#sqrt(x-4) = 2#

Square both sides (noting that this may introduce spurious solutions) to get:

#x-4 = 4#

Add #4# to both sides to get:

#x = 8#

Check that this works:

#sqrt(x+8)-sqrt(x-4) = sqrt(8+8)-sqrt(8-4) = sqrt(16)-sqrt(4) = 4-2 = 2#