basic square root question

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May 30th 2008, 07:00 PM
cmf0106

basic square root question

The expression
$\sqrt{3} * \sqrt[-4]{3}$

1.can be multiplied because
Expressions with radicals can be multiplied or divided as long as the root power or value under the radical is the same.

resulting in $\sqrt[-4]{9}$ ?

2. and just for fun how in the world would you divide

$\sqrt{3} / \sqrt[-4]{3}$ ?
May 30th 2008, 07:06 PM
TheEmptySet

Quote:

Originally Posted by cmf0106

The expression
$\sqrt{3} * \sqrt[-4]{3}$

1.can be multiplied because
Expressions with radicals can be multiplied or divided as long as the root power or value under the radical is the same.

resulting in $\sqrt[-4]{9}$ ?

2. and just for fun how in the world would you divide

$\sqrt{3} / \sqrt[-4]{3}$ ?

I'm sorry but that is not correct.

Remember that we can rewrite radicals as fractional exponents

$\sqrt[n]{a^m}=a^{\frac{m}{n}}$
What we should get is...

$3^{\frac{1}{2}}\cdot 3^{-\frac{1}{4}}=3^{\frac{1}{2}-\frac{1}{4}}=3^{\frac{1}{4}}=\sqrt[4]{3}$

Try this on the second one...
May 30th 2008, 08:20 PM
cmf0106

And just to clarify

$3^{\frac{1}{2}}\cdot 3^{-\frac{1}{4}}=3^{\frac{1}{2}-\frac{1}{4}}=3^{\frac{1}{4}}=\sqrt[4]{3}<br />$
the "-4" in $\sqrt[-4]{3}$ gets dropped when its turned into a fraction thus becoming 1/4, yes?
May 30th 2008, 08:54 PM
TheEmptySet

Quote:

Originally Posted by cmf0106

And just to clarify

$3^{\frac{1}{2}}\cdot 3^{-\frac{1}{4}}=3^{\frac{1}{2}-\frac{1}{4}}=3^{\frac{1}{4}}=\sqrt[4]{3}<br />$
the "-4" in $\sqrt[-4]{3}$ gets dropped when its turned into a fraction thus becoming 1/4, yes?

It didn't get dropped It just got moved

$\frac{1}{-4}=-\frac{1}{4}$

Then $\frac{1}{2}-\frac{1}{4}=\frac{2}{4}-\frac{1}{4}=\frac{1}{4}$

Good luck.
May 31st 2008, 06:37 AM
cmf0106

So then had the problem been

$\sqrt{3} / \sqrt[-4]{3}$

the answer would look like this?

$\sqrt{3} \frac{1}{2} - \frac{-1}{4} = \frac{3}{4}$

Thus the final answer would be $\sqrt[4]{3^3}$ ?
May 31st 2008, 07:05 AM
TheEmptySet

$\frac{\sqrt{3}}{\sqrt[-4]{3}}=\frac{3^{\frac{1}{2}}}{3^{-\frac{1}{4}}}=3^{\frac{1}{2}-\left( -\frac{1}{4}\right)}=3^\frac{3}{4}=\sqrt[4]{3^3}$

Yes that is correct.
May 31st 2008, 07:40 AM
cmf0106

And lastly these two examples, they will both have the same radicand and index.

Do these look correct as well?

1. $\sqrt[5]{3} * \sqrt[5]{3} = \frac{1}{5} + \frac{1}{5} = \sqrt[5]{9^2}$

2. $\sqrt[5]{3} / \sqrt[5]{3} = \frac{1}{5} - \frac{1}{5} = \sqrt{9}$ If this one is correct it would seem that the number will always be just a regular $\sqrt{x}$ . Because to divide the radicands and index's have to be same. Since the index's must be the same when you subtract one from the other it will result in 0.
May 31st 2008, 07:46 AM
TheEmptySet

Quote:

Originally Posted by cmf0106

And lastly these two examples, they will both have the same radicand and index.

2. $\sqrt[5]{3} / \sqrt[5]{3} = \frac{1}{5} - \frac{1}{5} = \sqrt{9}$ If this one is correct it would seem that the number will always be just a regular $\sqrt{x}$ . Because to divide the radicands and index's have to be same. Since the index's must be the same when you subtract one from the other it will result in 0.

Remember anything divided by itself is eqal to 1... and $x^0=1 \mbox{ if } x \ne 0$

$\frac{\sqrt[5]{3}}{\sqrt[5]{3}}=\frac{3^{\frac{1}{5}}}{3^{\frac{1}{5}}}=3^{\f rac{1}{5}-\frac{1}{5}}=3^0=1$
May 31st 2008, 07:54 AM
Jonboy

And if you're wondering why anything to the zero power beside 0 is 1, consider this.

Anything divided by itself is 1, right?

So if you have: $\frac{a^{b}}{a^{b}} = a^{b - b} = a^0$ .

Since $\frac{a^{b}}{a^{b}}$ is one, you can sub one in for it in the former equation to get $a^0 = 1$ .