# Simplifying radical expressions with fractions

• Oct 14th 2011, 04:57 PM
vaironxxrd
I'm having a tough time with the rules of simplifying radicals with fractions.

I don't understand why can't I reduce, $\displaystyle -\sqrt\frac{60}{5}$

to $\displaystyle -\sqrt\frac{12}{1}$
• Oct 14th 2011, 05:01 PM
skeeter
Re: Simplifying radical expressions with fractions
Quote:

Originally Posted by vaironxxrd
I'm having a tough time with the rules of simplifying radicals with fractions.

I don't understand why can't I reduce, $\displaystyle -\sqrt\frac{60}{5}$

to $\displaystyle -\sqrt\frac{12}{1}$

who said you cannot? $\displaystyle -\sqrt{\frac{60}{5}} = -\sqrt{12} = -2\sqrt{3}$
• Oct 14th 2011, 05:22 PM
vaironxxrd
Re: Simplifying radical expressions with fractions
Quote:

Originally Posted by skeeter
who said you cannot? $\displaystyle -\sqrt{\frac{60}{5}} = -\sqrt{12} = -2\sqrt{3}$

I got to stop my habit of not checking my answers. Sorry for taking of your time skeeter.
• Oct 15th 2011, 06:14 AM
vaironxxrd
Re: Simplifying radical expressions with fractions
Quote:

Originally Posted by skeeter
who said you cannot? $\displaystyle -\sqrt{\frac{60}{5}} = -\sqrt{12} = -2\sqrt{3}$

Skeeter if I have $\displaystyle -\frac{9}{2\sqrt45}$

Would the solution be:

$\displaystyle -\frac{9}{2\sqrt{3 \cdot 3 \cdot 5}}$
=

$\displaystyle -\frac{9}{2 \cdot 3 \sqrt5}$
=

$\displaystyle -\frac{9}{6\sqrt5}$

$\displaystyle -\frac{3}{2\sqrt5}$

$\displaystyle -\frac{3\sqrt5}{10}$
?
• Oct 15th 2011, 06:21 AM
e^(i*pi)
Re: Simplifying radical expressions with fractions
Quote:

Originally Posted by vaironxxrd
Skeeter if I have $\displaystyle -\frac{9}{2\sqrt45}$

Would the solution be:

$\displaystyle -\frac{9}{2\sqrt{3 \cdot 3 \cdot 5}}$
=

$\displaystyle -\frac{9}{2 \cdot 3 \sqrt5}$
=

$\displaystyle -\frac{9}{6\sqrt5}$

$\displaystyle -\frac{3}{2\sqrt5}$

$\displaystyle -\frac{3\sqrt5}{10}$
?

Yes, that's correct