How do you simplify $\sqrt{\frac{2}{3}}$ $sqrt(2/3)$ ?

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How do you simplify $\sqrt{\frac{2}{3}}$ $sqrt(2/3)$ ?

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Answer 1 · 2017-04-21T05:19:41

$\sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}} (\frac{\sqrt{3}}{\sqrt{3}}) = \frac{\sqrt{2} \sqrt{3}}{\sqrt{3} \sqrt{3}} = \frac{\sqrt{6}}{3}$ $sqrt(2/3)=sqrt2/sqrt3(sqrt3/sqrt3)=(sqrt2sqrt3)/(sqrt3sqrt3)=sqrt6/3$

Explanation:

$\sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}}$ $sqrt(2/3)=sqrt2/sqrt3$

and now we can see that there is a square root in the denominator that doesn't belong. So let's get rid of that.

$\frac{\sqrt{2}}{\sqrt{3}} (\frac{\sqrt{3}}{\sqrt{3}}) = \frac{\sqrt{2} \sqrt{3}}{\sqrt{3} \sqrt{3}} = \frac{\sqrt{6}}{3}$ $sqrt2/sqrt3(sqrt3/sqrt3)=(sqrt2sqrt3)/(sqrt3sqrt3)=sqrt6/3$

Answer 2 · 2017-04-21T06:08:33

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

Explanation:

Note that if $a \geq 0$ $a >= 0$ and $b > 0$ $b > 0$ then:

$\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$ $sqrt(a/b) = sqrt(a)/sqrt(b)$

Also if $a \geq 0$ $a >= 0$ then:

$\sqrt{a^{2}} = a$ $sqrt(a^2) = a$

Instead of breaking up the square root then rationalising the denominator, we can make the denominator square first as follows:

$\sqrt{\frac{2}{3}} = \sqrt{\frac{2 \cdot 3}{3 \cdot 3}} = \sqrt{\frac{6}{3^{2}}} = \frac{\sqrt{6}}{\sqrt{3^{2}}} = \frac{\sqrt{6}}{3}$ $sqrt(2/3) = sqrt((2*3)/(3*3)) = sqrt(6/3^2) = sqrt(6)/sqrt(3^2) = sqrt(6)/3$

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How do you simplify $\sqrt{\frac{2}{3}}$ $sqrt(2/3)$ ?

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