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

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Answer 1 · 2016-08-19T15:45:19

#x=1/2#

Lets get rid of the square root on the left hand side (LHS)

Square both sides giving:

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

#cancel(2x)+3=9-6sqrt(2x)+cancel(2x)#

#sqrt(2x)=(9-3)/6 = +1#

square both sides again

#2x=1 => x=1/2#