# How do you simplify sqrt735/sqrt5?

Jun 30, 2016

$\frac{\sqrt{735}}{\sqrt{5}} = 7 \sqrt{3}$

#### Explanation:

$\frac{\sqrt{735}}{\sqrt{5}}$

= $\frac{\sqrt{3 \times 5 \times 7 \times 7}}{\sqrt{5}}$

= $\sqrt{\frac{3 \times 5 \times 7 \times 7}{5}}$

= $\sqrt{\frac{3 \times \cancel{5} \times 7 \times 7}{\cancel{5}}}$

= $\sqrt{3 \times \underline{7 \times 7}}$

= $7 \sqrt{3}$

Jul 4, 2016

$7 \sqrt{3}$

#### Explanation:

$\textcolor{b l u e}{\text{Method}}$

We need to look for integer factors of 735 and with a bit of luck be able to cancel out the $\sqrt{5}$ denominator.

Suppose we had 2 unknown variables $a \text{ and } b$. Suppose these 2 variables were presented in the form of

$\frac{\sqrt{a}}{\sqrt{b}}$ This we can write as $\sqrt{\frac{a}{b}}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Answering your question}}$

Notice that the sum of the digits in 735 is 15. As 15 is divisible by 3 then 735 is also divisible by 3

From the factor tree we observe that 735 is the product of $3 \times 5 \times {7}^{2}$

Write $\frac{\sqrt{735}}{\sqrt{5}}$ as $\sqrt{\frac{3 \times \cancel{5} \times {7}^{2}}{\cancel{5}}}$

We can take the ${7}^{2}$ outside the square root but it becomes just $7$ giving:

$7 \sqrt{3}$