# How do you rationalize the denominator and simplify (3sqrt9)/sqrt27?

Jul 12, 2018

$\sqrt{3}$

#### Explanation:

You can simplify the expression quite a lot before rationalizing:

• $\sqrt{9} = \sqrt{{3}^{2}} = 3$
• $\sqrt{27} = \sqrt{9 \cdot 3} = \sqrt{9} \cdot \sqrt{3} = 3 \sqrt{3}$

So, the expression becomes

$\setminus \frac{\cancel{3} \cdot 3}{\cancel{3} \sqrt{3}}$

To rationalize this expression, multiply and divide by $\sqrt{3}$:

$\setminus \frac{3}{\sqrt{3}} = \setminus \frac{3}{\sqrt{3}} \cdot \setminus \frac{\sqrt{3}}{\sqrt{3}} = \setminus \frac{3 \sqrt{3}}{\sqrt{3} \sqrt{3}} = \setminus \frac{\cancel{3} \sqrt{3}}{\cancel{3}} = \sqrt{3}$

Jul 12, 2018

$\sqrt{3}$

#### Explanation:

$\frac{3 \sqrt{9}}{\sqrt{27}} = \frac{3 \cdot 3}{3 \sqrt{3}} = \frac{3}{\sqrt{3}} = \frac{3 \sqrt{3}}{3} = \sqrt{3}$

Jul 12, 2018

$\sqrt{3}$

#### Explanation:

$\frac{3 \sqrt{9}}{\sqrt{27}}$

$\therefore = \frac{3 \sqrt{9}}{\sqrt{27}} \times \frac{\sqrt{27}}{\sqrt{27}}$

$\sqrt{27} \cdot \sqrt{27} = 27$

$\therefore = \frac{3 \sqrt{9 \cdot 27}}{27}$

$\therefore = \frac{3 \sqrt{\left(3 \cdot 3\right) \left(3 \cdot 3 \cdot 3\right)}}{27}$

$\sqrt{3} \cdot \sqrt{3} = 3$

$\therefore = \frac{3 \cdot 3 \cdot 3 \sqrt{3}}{27}$

$\therefore = \frac{{\cancel{27}}^{1} \sqrt{3}}{\cancel{27}} ^ 1$

$\therefore = \sqrt{3}$

Jul 12, 2018

$\sqrt{3}$

#### Explanation:

First, we can rewrite $\textcolor{b l u e}{\sqrt{27}}$ as $\sqrt{3} \cdot \sqrt{9}$, which is the same as $3 \sqrt{3}$. Doing this leaves us with

$\frac{3 \sqrt{9}}{\textcolor{b l u e}{3 \sqrt{3}}}$

A $3$ in the numerator and denominator cancels, leaving us with

$\frac{\sqrt{9}}{\sqrt{3}}$, which is also equal to

$\frac{3}{\sqrt{3}}$

The key realization is that when we rationalize the denominator, we multiply the top and bottom by that denominator.

In our example, we have

$\frac{3 \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}}$

Since we are multiplying by $1$ essentially, we are not changing the expression's value. The denominator simplifies to $3$, and we're left with

$\sqrt{3}$

Even though we could have rationalized the denominator in the intermediate steps, it never hurts to simplify first.

Hope this helps!