# How do you simplify  (2+sqrt(x)) / (sqrt(2 x)+sqrt(8)) ?

$\sqrt{2}$

#### Explanation:

When working a problem such as this, keep in mind that if we take something (like our denominator) that has the form of $a + b$, we can multiply by $a - b$ and we'll end up with ${a}^{2} - {b}^{2}$. And so:

$\frac{2 + \sqrt{x}}{\sqrt{2 x} + \sqrt{8}} \left(1\right)$

$\frac{2 + \sqrt{x}}{\sqrt{2 x} + \sqrt{8}} \left(\frac{\sqrt{2 x} - \sqrt{8}}{\sqrt{2 x} - \sqrt{8}}\right)$

$\frac{\left(2 + \sqrt{x}\right) \left(\sqrt{2 x} - \sqrt{8}\right)}{\left(\sqrt{2 x} + \sqrt{8}\right) \left(\sqrt{2 x} - \sqrt{8}\right)}$

$\frac{2 \sqrt{2 x} - 2 \sqrt{8} + \sqrt{x} \sqrt{2 x} - \sqrt{x} \sqrt{8}}{2 x - 8}$

Now for some simplification in the numerator. Remember that $\sqrt{8} = 2 \sqrt{2}$:

$\frac{2 \sqrt{2 x} - 2 \left(2 \sqrt{2}\right) + \sqrt{2 {x}^{2}} - 2 \sqrt{2 x}}{2 x - 8}$

$\frac{- 4 \sqrt{2} + x \sqrt{2}}{2 x - 8}$

$\frac{\sqrt{2} \left(x - 4\right)}{2 \left(x - 4\right)}$

$\frac{\sqrt{2}}{2}$

Remember that ${\left(\pm \sqrt{2}\right)}^{2} = 2$

$\frac{\sqrt{2}}{\sqrt{2}} ^ 2 = \sqrt{2}$

and

$\frac{\sqrt{2}}{- \sqrt{2}} ^ 2 = \sqrt{2}$

Jul 2, 2018

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

#### Explanation:

Given: $\frac{2 + \sqrt{x}}{\sqrt{2 x} + \sqrt{8}}$

Multiply both the numerator and the denominator by the conjugate of the denominator, which is basically multiplying the expression by $1$.

The conjugate is a factor that when multiplied, cancels the middle term. The conjugate of $\sqrt{2 x} + \sqrt{8}$ is $\sqrt{2 x} - \sqrt{8} .$

$\frac{2 + \sqrt{x}}{\sqrt{2 x} + \sqrt{8}} \cdot \frac{\sqrt{2 x} - \sqrt{8}}{\sqrt{2 x} - \sqrt{8}}$

$= \frac{2 \sqrt{2 x} - 2 \sqrt{8} + \sqrt{x} \sqrt{2 x} - \sqrt{x} \sqrt{8}}{\sqrt{2 x} \sqrt{2 x} - \sqrt{8} \sqrt{8}}$

Use the Radical rules: sqrt(a)sqrt(b) = sqrt(ab); " "sqrt(x)sqrt(x) = sqrt(x^2) = x

Simplify $\sqrt{8} : \text{ } \sqrt{8} = \sqrt{4 \cdot 2} = \sqrt{4} \sqrt{2} = 2 \sqrt{2}$

$= \frac{\cancel{2 \sqrt{2 x}} - 2 \cdot 2 \sqrt{2} + x \sqrt{2} \cancel{- 2 \sqrt{2 x}}}{\sqrt{4 {x}^{2}} - \sqrt{64}}$

$= \frac{- 4 \sqrt{2} + x \sqrt{2}}{2 x - 8} = \frac{x \sqrt{2} - 4 \sqrt{2}}{2 x - 8}$

Factor: $\text{ } \frac{\sqrt{2} \cancel{x - 4}}{2 \cancel{x - 4}}$

$\frac{2 + \sqrt{x}}{\sqrt{2 x} + \sqrt{8}} = \frac{\sqrt{2}}{2}$