# How do you simplify sqrt27*sqrt33?

Feb 6, 2016

9sqrt11

#### Explanation:

sqrt27 * sqrt33

= [sqrt39] * [sqrt311]
= [3sqrt3] * [sqrt311] ; sqrt27 becomes 3sqrt3 because the sqrt of 9 is 3
= [3sqrt3] * [sqrt3] * [sqrt11] ; we separate sqrt3 and sqrt11
= 3sqrt3
sqrt11 ; we multiply those with sqrt3
= 3sqrt9 * sqrt11
= 3 * 3 * sqrt11 ; there are two 3s now because again the sqrt of 9 is 3
= 9sqrt11 ; FINAL ANSWER

Feb 6, 2016

First, before multiplying, you can simplify the √27.

#### Explanation:

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

= $3 \sqrt{3}$

Now we can multiply, multiplying radicals with radicals and whole numbers with whole numbers.

$3 \sqrt{3} \times \sqrt{33}$

= $3 \sqrt{99}$

= $3 \sqrt{9 \times 11}$

= $3 \left(3\right) \sqrt{11}$

= $9 \sqrt{11}$

So, $9 \sqrt{11}$ is your answer, in simplest form.

Practice exercises:

1. Simplify:

a) $3 \sqrt{5} \times 4 \sqrt{7}$

b) $\sqrt{24} \times 2 \sqrt{48}$

2 . Solve for x in $\sqrt{6 x} \times \sqrt{2} = 2 x$

Good Luck!