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

Jun 19, 2016

$\sqrt{2 x + 3} = 2 + \sqrt{x + 2}$

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

$2 x + 3 = 4 + 4 \sqrt{x + 2} + x + 2$

$2 x + 3 - 4 - x - 2 = 4 \sqrt{x + 2}$

$x - 3 = 4 \sqrt{x + 2}$

${\left(x - 3\right)}^{2} = {\left(4 \sqrt{x + 2}\right)}^{2}$

${x}^{2} - 6 x + 9 = 16 \left(x + 2\right)$

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

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

$\left(x - 23\right) \left(x + 1\right) = 0$

$x = 23 \mathmr{and} x = - 1$

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

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

Hopefully this helps!