# What is the simplification of sqrt((18x^5y^4)/(49xz^3))?

Jul 31, 2016

$\frac{3 {x}^{2} {y}^{2} \sqrt{2 z}}{7 {z}^{2}}$

#### Explanation:

$\textcolor{b l u e}{\text{General comment}}$

Any values that are squares can be taken outside the square root. So for example if you had $\sqrt{{2}^{2} \times 3}$ you could write this as
$2 \sqrt{3}$. Or if you had $\sqrt{{2}^{4} \times 3}$ you could write ${2}^{2} \sqrt{3}$

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$\textcolor{b l u e}{\text{Answering the question}}$

Write as sqrt(2xx3^2xx x xx x^4xx y^4)/(sqrt(7^2 xz xx z^2) giving:

$\frac{3 {x}^{2} {y}^{2} \sqrt{2 x}}{7 z \sqrt{x z}}$

Write as:$\text{ } \frac{3 {x}^{2} {y}^{2} \sqrt{2} \cancel{\sqrt{x}}}{7 z \sqrt{z} \cancel{\sqrt{x}}}$

Multiply by 1 but in the form of $1 = \frac{\sqrt{z}}{\sqrt{z}}$ giving

$\frac{3 {x}^{2} {y}^{2}}{7 z} \times \frac{\sqrt{2}}{\sqrt{z}} \times \frac{\sqrt{z}}{\sqrt{z}}$

$\frac{3 {x}^{2} {y}^{2} \sqrt{2 z}}{7 {z}^{2}}$