# How do you simplify sqrt(x^11)+sqrt(x^5)?

May 24, 2018

See a solution process below:

#### Explanation:

First, we can rewrite this as:

$\sqrt{{x}^{10} \cdot x} + \sqrt{{x}^{4} \cdot x}$

Next, we can use this rule for radicals to simplify each of the terms:

$\sqrt{\textcolor{red}{a} \cdot \textcolor{b l u e}{b}} = \sqrt{\textcolor{red}{a}} \cdot \sqrt{\textcolor{b l u e}{b}}$

$\sqrt{\textcolor{red}{{x}^{10}} \cdot \textcolor{b l u e}{x}} + \sqrt{\textcolor{red}{{x}^{4}} \cdot \textcolor{b l u e}{x}} \implies$

$\left(\sqrt{\textcolor{red}{{x}^{10}}} \cdot \sqrt{\textcolor{b l u e}{x}}\right) + \left(\sqrt{\textcolor{red}{{x}^{4}}} \cdot \sqrt{\textcolor{b l u e}{x}}\right) \implies$

${x}^{5} \sqrt{\textcolor{b l u e}{x}} + {x}^{2} \sqrt{\textcolor{b l u e}{x}}$

We can now factor out the common term giving:

$\left({x}^{5} + {x}^{2}\right) \sqrt{\textcolor{b l u e}{x}}$

If necessary, we can also factor out a common term from the two terms within the parenthesis:

$\left({x}^{3} {x}^{2} + 1 {x}^{2}\right) \sqrt{\textcolor{b l u e}{x}}$

$\left({x}^{3} + 1\right) {x}^{2} \sqrt{\textcolor{b l u e}{x}}$

I think the correct version is

$\sqrt{{x}^{11}} + \sqrt{{x}^{5}}$

$\sqrt{{\left({x}^{5}\right)}^{2} \cdot x} + \sqrt{{\left({x}^{2}\right)}^{2} \cdot x}$

$| {x}^{5} | \cdot \sqrt{x} + {x}^{2} \cdot \sqrt{x}$

${x}^{4} \cdot | x | \cdot \sqrt{x} + {x}^{2} \cdot \sqrt{x}$

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

where || =absolute value