# How do you simplify (sqrt(5k^4)+3sqrt(2k))/sqrt(3k^3)?

Jul 19, 2017

See a solution process belowL

#### Explanation:

First, we can rationalize the denominator by multiplying the expression by the appropriate form or $1$ to remove the radical from the denominator while keeping the value of the expression the same:

$\frac{\sqrt{5 {k}^{4}} + 3 \sqrt{2 k}}{\sqrt{3 {k}^{3}}} \implies \frac{\sqrt{3 {k}^{3}}}{\sqrt{3 {k}^{3}}} \times \frac{\sqrt{5 {k}^{4}} + 3 \sqrt{2 k}}{\sqrt{3 {k}^{3}}} \implies$

$\frac{\sqrt{3 {k}^{3}} \left(\sqrt{5 {k}^{4}} + 3 \sqrt{2 k}\right)}{\sqrt{3 {k}^{3}} \sqrt{3 {k}^{3}}} \implies$

$\frac{\sqrt{3 {k}^{3}} \sqrt{5 {k}^{4}} + 3 \sqrt{3 {k}^{3}} \sqrt{2 k}}{\sqrt{3 {k}^{3}}} ^ 2 \implies$

$\frac{\sqrt{3 {k}^{3}} \sqrt{5 {k}^{4}} + 3 \sqrt{3 {k}^{3}} \sqrt{2 k}}{3 {k}^{3}}$

We can simplify the numerator using this rule for radicals:

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

$\frac{\sqrt{3 {k}^{3}} \sqrt{5 {k}^{4}} + 3 \sqrt{3 {k}^{3}} \sqrt{2 k}}{3 {k}^{3}} \implies$

$\frac{\sqrt{3 {k}^{3} \cdot 5 {k}^{4}} + 3 \sqrt{3 {k}^{3} \cdot 2 k}}{3 {k}^{3}} \implies$

$\frac{\sqrt{15 {k}^{7}} + 3 \sqrt{6 {k}^{4}}}{3 {k}^{3}}$

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

$\frac{\sqrt{15 {k}^{7}} + 3 \sqrt{6 {k}^{4}}}{3 {k}^{3}} \implies \frac{\sqrt{{k}^{6} \cdot 15 k} + 3 \sqrt{{k}^{4} \cdot 6}}{3 {k}^{3}} \implies$

$\frac{{k}^{3} \sqrt{15 k} + 3 {k}^{2} \sqrt{6}}{3 {k}^{3}}$

If necessary, we can simplify further as:

$\frac{{k}^{3} \sqrt{15 k} + 3 {k}^{2} \sqrt{6}}{3 {k}^{3}} \implies \frac{{k}^{3} \sqrt{15 k}}{3 {k}^{3}} + \frac{3 {k}^{2} \sqrt{6}}{3 {k}^{3}} \implies$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{{k}^{3}}}} \sqrt{15 k}}{3 \textcolor{red}{\cancel{\textcolor{b l a c k}{{k}^{3}}}}} + \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} {k}^{2} \sqrt{6}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} {k}^{3}} \implies \frac{\sqrt{15 k}}{3} + \frac{{k}^{2} \sqrt{6}}{k} ^ 3 \implies$

$\frac{\sqrt{15 k}}{3} + \frac{\sqrt{6}}{k}$