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

Apr 8, 2018

$x = 8$

#### Explanation:

$\sqrt{x + 1} + 1 = \sqrt{2 x} , x > 0$

$1 = \sqrt{x + 1} - \sqrt{2 x}$ subtract $\sqrt{x - 1}$ from both sides

$1 = 2 x - 2 \sqrt{2 x \left(x + 1\right)} + \left(x + 1\right)$ square both sides

$0 = 3 x - 2 \sqrt{2 x \left(x + 1\right)}$ collect terms and subtract $1$ from both sides

$\sqrt{2 x \left(x + 1\right)} = \frac{3 x}{2}$ add $\sqrt{2 x \left(x + 1\right)}$ to both sides

$2 x \left(x + 1\right) = \frac{9 {x}^{2}}{4}$ square both sides

$2 {x}^{2} + 2 x = \frac{9 {x}^{2}}{4}$ expand brackets

$0 = {x}^{2} / 4 - 2 x$ move terms to RHS

$0 = x \left(\frac{x}{4} - 2\right)$ factorise

$x = 0 \mathmr{and} x = 8$ solve linear equations

Since $x > 0$, $x = 8$

Apr 8, 2018

$x = 8$

#### Explanation:

Given:

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

Square both sides (noting that squaring can introduce extraneous solutions) to get:

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

Subtract $x + 2$ from both sides to get:

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

Square both sides to get:

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

Subtract $4 x + 4$ from both sides to get:

$0 = {x}^{2} - 8 x = x \left(x - 8\right)$

So $x = 0$ or $x = 8$

Check whether these are solutions of the original equation:

$\sqrt{\left(\textcolor{b l u e}{0}\right) + 1} + 1 = 2 \ne 0 = \sqrt{2 \left(\textcolor{b l u e}{0}\right)}$

$\sqrt{\left(\textcolor{b l u e}{8}\right) + 1} + 1 = 3 + 1 = 4 = \sqrt{16} = \sqrt{2 \left(\textcolor{b l u e}{8}\right)}$

So $x = 8$ is the only solution of the given equation.