How do you solve $sqrt(2x+3)-sqrt(x+2)=2$ and find any extraneous solutions?

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Answer 1 · 2016-06-19T15:55:26

$sqrt(2x + 3) = 2 + sqrt(x + 2)$

$(sqrt(2x + 3))^2 = (2 + sqrt(x + 2))^2$

$2x + 3 = 4 + 4sqrt(x + 2) + x+ 2$

$2x + 3 - 4 - x - 2 = 4sqrt(x + 2)$

$x - 3 = 4sqrt(x + 2)$

$(x - 3)^2 = (4sqrt(x + 2))^2$

$x^2 - 6x + 9 = 16(x + 2)$

$x^2 - 6x + 9 = 16x + 32$

$x^2 - 22x -23 = 0$

$(x - 23)(x + 1) = 0$

$x = 23 and x = -1$

Checking in our original equation, we find that $x = 23$ works but $x = -1$ does not.

Our solution set is therefore ${x = 23}$ .

Hopefully this helps!

How do you solve sqrt(2x+3)-sqrt(x+2)=2sqrt(2x+3)-sqrt(x+2)=2 and find any extraneous solutions?