# How would you simplify sqrt(2x+3) - sqrt(x+1 )=1?

Feb 9, 2016

You must mean how to solve. I'll solve the equation for you.

#### Explanation:

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

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

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

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

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

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

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

${x}^{2} + 2 x + 1 = 4 \left(x + 1\right)$

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

${x}^{2} + 2 x - 4 x + 1 - 4 = 0$

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

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

$x = 3 \mathmr{and} x = - 1$

Always check the solutions in the original equation to make sure they aren't extraneous. If they do not work in the original equation, you must reject them.

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

So, x = 3 works. Now, let's check x = -1:

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

So, x = -1 works as well.

Your solution set would be x = 3, -1

Practice exercises:

1. Solve for x.

a) $\sqrt{3 x - 2} - \sqrt{x - 2} = 2$

b) $\sqrt{4 x + 5} + \sqrt{8 x + 9} = 12$