# How do you simplify 4 sq.root of 75 + sq.root of 27.?

Jul 13, 2015

$4 \sqrt{75} + \sqrt{27} = 23 \sqrt{3}$

#### Explanation:

We are given:

$4 \sqrt{75} + \sqrt{27}$

Let's look at $\sqrt{75}$

$25$ goes into $75$ $3$ times

$3 \cdot 25 = 75$

So,

$\sqrt{75} = \sqrt{3 \cdot 25}$

the square root of $25$ is $5$, so

$\sqrt{75} = \sqrt{3 \cdot 25} = 5 \sqrt{3}$

Now, let's look at $\sqrt{27}$

$3$ goes into $27$ $9$ times

$3 \cdot 9 = 27$

So,

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

the square root of $9$ is $3$, so

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

So,

$4 \sqrt{75} + \sqrt{27} = 4 \cdot 5 \sqrt{3} + 3 \sqrt{3}$

$4 \cdot 5 \sqrt{3} + 3 \sqrt{3} = 20 \sqrt{3} + 3 \sqrt{3} = 23 \sqrt{3}$

Recall: $a \sqrt{b} + c \sqrt{b} = \left(a + c\right) \sqrt{b}$

Jul 13, 2015

$4 \sqrt{75} + \sqrt{27} = 23 \sqrt{3}$

#### Explanation:

Use $\sqrt{a b} = \sqrt{a} \sqrt{b}$ for $a , b \ge 0$

$4 \sqrt{75} + \sqrt{27}$

$= 4 \sqrt{{5}^{2} \cdot 3} + \sqrt{{3}^{2} \cdot 3}$

$= 4 \sqrt{{5}^{2}} \sqrt{3} + \sqrt{{3}^{2}} \sqrt{3}$

$= 4 \cdot 5 \sqrt{3} + 3 \sqrt{3}$

$= 20 \sqrt{3} + 3 \sqrt{3}$

$= 23 \sqrt{3}$