How do you solve $\sqrt{2 x + 3} = 3 - \sqrt{2 x}$ $sqrt(2x+3)=3-sqrt(2x)$ ?

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Answer 1 · 2016-08-19T15:45:19

$x = \frac{1}{2}$ $x=1/2$

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

Square both sides giving:

$2 x + 3 = {(3 - \sqrt{2 x})}^{2}$ $2x+3=(3-sqrt(2x))^2$

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

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

square both sides again

$2x=1 => x=1/2$

How do you solve sqrt(2x+3)=3-sqrt(2x)√2x+3=3−√2xsqrt(2x+3)=3-sqrt(2x)?