# How do you rationalize the dominator and simplify for the square root of 35 over the square root of 55?

Nov 5, 2015

$\frac{\sqrt{77}}{11}$

#### Explanation:

Your expression is $\frac{\sqrt{35}}{\sqrt{55}}$. Since you can always mutiply an expression by $1$ without changing its value, and you can see $1$ as any number divided by itself, we have

$\frac{\sqrt{35}}{\sqrt{55}} = \frac{\sqrt{35}}{\sqrt{55}} \cdot 1 = \frac{\sqrt{35}}{\sqrt{55}} \cdot \frac{\sqrt{55}}{\sqrt{55}}$

Doing the multiplications (remember that $\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}$), we have

$\frac{\sqrt{35 \cdot 55}}{\sqrt{{55}^{2}}}$

And of course $\sqrt{{a}^{2}} = a$ (if $a$ is positive), so we have

$\frac{\sqrt{1925}}{55}$.

Finally, we can factor $1925$ with prime numbers, and we have

$1925 = {5}^{2} \cdot 7 \cdot 11$, and so $\sqrt{1925} = \sqrt{{5}^{2} \cdot 7 \cdot 11} = 5 \sqrt{7 \cdot 11}$, and $5$ and $55$ cancel out.