dividing square roots

**imbadatmath** · April 20th 2008, 05:26 PM

$\sqrt{35x}/\sqrt{7x}$

**imbadatmath** · April 20th 2008, 05:28 PM

was not sure how to write it but the the square root of 35x on top and the square root of 7x on the bottem

**Mathstud28** · April 20th 2008, 05:38 PM

Originally Posted by imbadatmath

$\sqrt{35x}/\sqrt{7x}$

how about I give you a BIG hint...this can be rewritten as $\bigg(\frac{35x}{7x}\bigg)^{\frac{1}{2}}$

**imbadatmath** · April 20th 2008, 06:41 PM

im still confused lol

**imbadatmath** · April 20th 2008, 07:49 PM

y would i make it into a multuplication problem?

**o_O** · April 20th 2008, 08:42 PM

It's not a multiplication problem ... Basically, MathStud put the two square roots into one:

$\frac{\sqrt{35x}}{\sqrt{7x}} = \sqrt{\frac{35x}{7x}} = \text{Etc.}$

since as he said:
$\frac{a^{c}}{b^{c}} = \left(\frac{a}{b}\right)^{c}$

where $c = \frac{1}{2}$ here

**imbadatmath** · April 21st 2008, 02:37 AM

so its basically 35/7 equals 5? is that the answer?

**topsquark** · April 21st 2008, 03:13 AM

Originally Posted by imbadatmath

so its basically 35/7 equals 5? is that the answer?

$\sqrt{\frac{35}{7}} = \sqrt{5}$
If that's what you meant, then yes.

-Dan

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