# How do you solve sqrt(2x+3)-sqrt( x+1) =1?

Apr 29, 2018

The two solutions are $x = 3$ and $x = - 1$.

#### Explanation:

Isolate one of the radicals, square both sides, isolate the other radical, then square both sides again:

$\sqrt{2 x + 3} - \sqrt{x + 1} = 1$

$\sqrt{2 x + 3} = 1 + \sqrt{x + 1}$

${\left(\sqrt{2 x + 3}\right)}^{2} = {\left(1 + \sqrt{x + 1}\right)}^{2}$

$2 x + 3 = 1 + 2 \sqrt{x + 1} + x + 1$

$x + 1 = 2 \sqrt{x + 1}$

$\frac{x + 1}{2} = \sqrt{x + 1}$

${\left(x + 1\right)}^{2} / {2}^{2} = {\left(\sqrt{x + 1}\right)}^{2}$

$\frac{{x}^{2} + 2 x + 1}{4} = x + 1$

${x}^{2} + 2 x + 1 = 4 x + 4$

${x}^{2} - 2 x - 3 = 0$

$\left(x - 3\right) \left(x + 1\right) = 0$

$x = 3 , - 1$

Those are the solutions to the problem. Hope this helped!