How do you simplify $\sqrt[4]{32}$ $root4 32$ ?

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How do you simplify $\sqrt[4]{32}$ $root4 32$ ?

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Answer 1 · 2016-03-08T16:01:37

$32 = 16 \cdot 2$ $32 = 16 * 2$
$32^{\frac{1}{4}} = 16^{\frac{1}{4}} \cdot 2^{\frac{1}{4}}$ $32^(1/4) = 16^(1/4) * 2^(1/4)$
$= 2 \cdot 2^{\frac{1}{4}}$ $= 2 * 2^(1/4)$

Answer 2 · 2016-03-08T16:04:20

$\sqrt[4]{32} = 2^{5/4} = 2 \sqrt[4]{2}$ $root4 32=2^("5/4")=2root4 2$

Explanation:

We should first note that $32 = 2^{5}$ $32=2^5$ .

$= \sqrt[4]{2^{5}}$ $=root4(2^5)$

To simplify this, use the rule:

$\sqrt[b]{x^{a}} = x^{a / b}$ $rootb(x^a)=x^(a"/"b)$

Thus, the expression is equal to

$= 2^{5/4}$ $=2^("5/4")$

This is simplified. However, if you wish, there is another way to simplify it:

$= 2^{4 / 4 + 1/4}$ $=2^(4"/"4+"1/4")$

$= 2^{1 + 1/4}$ $=2^(1+"1/4")$

Use the rule:

$x^{a + b} = x^{a} (x^{b})$ $x^(a+b)=x^a(x^b)$

So the expression can be split up into:

$= 2^{1} (2^{\frac{1}{4}})$ $=2^1(2^(1/4))$

Which equals

$= 2 \sqrt[4]{2}$ $=2root4 2$

Another way of approaching this problem is to say that $\sqrt[4]{32}$ $root4 32$ equals

$= \sqrt[4]{2^{4} \cdot 2}$ $=root4(2^4*2)$

The $2^{4}$ $2^4$ can be brought out of the root:

$= 2 \sqrt[4]{2}$ $=2root4 2$

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