How do you simplify $\frac{\sqrt{8}}{\sqrt{3}}$ $sqrt 8/ sqrt3$ ?

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How do you simplify $\frac{\sqrt{8}}{\sqrt{3}}$ $sqrt 8/ sqrt3$ ?

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Answer 1 · 2018-03-28T20:55:25

To answer is this: $\frac{\sqrt{8}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{8} \cdot \sqrt{3}}{3} = \frac{\sqrt{24}}{3}$ $sqrt(8)/(sqrt3)*sqrt(3)/(sqrt(3))=(sqrt8*sqrt3)/3=(sqrt24)/3$

Explanation:

In this question, since you do not want square root functions as the divisors, you multiply the top and bottom by the square root divisor at the bottom.

In this case, $\sqrt{3}$ $sqrt(3)$ . After multiplying the top and bottom by $\sqrt{3}$ $sqrt(3)$ , you will remove the square root term from the bottom and as such getting you only $3$ $3$ , but the top will be $\sqrt{8} \cdot \sqrt{3}$ $sqrt 8 * sqrt 3$ , which is $\sqrt{24}$ $sqrt 24$ .

In your final answer, it will then become $\frac{\sqrt{24}}{3}$ $sqrt(24)/ 3$ .

Answer 2 · 2018-03-28T20:56:48

$\frac{2 \sqrt{6}}{3}$ $(2sqrt6)/3$

Explanation:

rationalise the denominator, by multiplying by $\sqrt{3} :$ $sqrt3:$

$\frac{\sqrt{3} \cdot \sqrt{8}}{\sqrt{3} \cdot \sqrt{3}}$ $(sqrt3 * sqrt8)/(sqrt3 * sqrt3)$

$= \frac{\sqrt{24}}{3}$ $= (sqrt24)/3$

$\sqrt{24} = \sqrt{4} \cdot \sqrt{6}$ $sqrt24 = sqrt4 * sqrt6$

$= 2 \cdot \sqrt{6} = 2 \sqrt{6}$ $= 2 * sqrt6 = 2sqrt6$

$\frac{\sqrt{24}}{3} = \frac{2 \sqrt{6}}{3}$ $(sqrt24)/3 = (2sqrt6)/3$

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