Change/simplify the index of a radical

**jesse** · May 20th 2011, 02:33 PM

$\sqrt[4]{64}-5\sqrt[6]{\frac{1}{8}} = 2\sqrt{2}-\frac{5}{2}\sqrt{2} = -\frac{1}{2}\sqrt{2}$

I'm not sure how the above all works out. I can't find a tutorial online for simplifying the index of a radical. Can someone help me out with an explanation?

**pickslides** · May 20th 2011, 02:57 PM

Lets start with the first term,

$\displaystyle \sqrt[4]{64} = \sqrt[4]{2^6} = 2^{\frac{6}{4}} = 2^{\frac{3}{2}} = \sqrt{2^3} = \sqrt{8} = \sqrt{2\times 4} = \sqrt{2}\sqrt{4} = 2\sqrt{2}$

There are other ways to reach this result, you just need additional exposure to the index laws.

Do you follow?

**jesse** · May 20th 2011, 03:05 PM

I get it. I didn't think to put a power on the radicand. Thanks for your help!

**Plato** · May 20th 2011, 03:09 PM

Originally Posted by jesse

$\sqrt[4]{64}-5\sqrt[6]{\frac{1}{8}} = 2\sqrt{2}-\frac{5}{2}\sqrt{2} = -\frac{1}{2}\sqrt{2}$

It is important to understand these equivalent statements.
$\sqrt[4]{64}=\left(2^{\frac{6}{4}}\right)=\left(2^{\frac{3 }{2}}\right)=2\sqrt{2}$

And $\sqrt[6]{\frac{1}{8}}=\left(2^{-\frac{3}{6}}\right)$

Thread: Change/simplify the index of a radical

LinkBack

Thread Tools

Display

Change/simplify the index of a radical

Similar Math Help Forum Discussions

Index change in Double summation

Simplify using index laws.

simplify radical

simplify radical

simplify radical