Simplifying Radicals: 1/5 inside a square root symbol

jramirez23

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Jan 23, 2008
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I have a problem with this following problem where I have to simplify a radical expression.

The expression is 1/5 inside a square root symbol.

I know that I have to distribute the square root symbol individually to both the numerator and denominator; so the expression will become a 1 inside a square root symbol OVER 5 inside a square root symbol.

This is where I really become confused. I can't find a way to eliminate the radicand from the denominator so it will simplify.
 
Re: Simplifying Radicals

jramirez23 said:
I have a problem with this following problem where I have to simplify a radical expression.

The expression is 1/5 inside a square root symbol.

I know that I have to distribute the square root symbol individually to both the numerator and denominator; so the expression will become a 1 inside a square root symbol OVER 5 inside a square root symbol.

This is where I really become confused. I can't find a way to eliminate the radicand from the denominator so it will simplify.

\(\displaystyle \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}}\) or if you prefer, you can multiply top and bottom by \(\displaystyle \sqrt{5}\) to give you \(\displaystyle \frac{\sqrt{5}}{5}\)

Either way is acceptable and equal.
 
Re: Simplifying Radicals

I'm having another problem, but I get what you have told me.

This new problem that I can't solve is the following:

4 inside a square root symbol
OVER
10 inside a square root symbol

What I did was multiply by the square root of 10 over itself to the original equation.

I then got 40 inside of a square root symbol over 10. This is obviously wrong.
 
Re: Simplifying Radicals

jramirez23 said:
I'm having another problem, but I get what you have told me.

This new problem that I can't solve is the following:

4 inside a square root symbol
OVER
10 inside a square root symbol

What I did was multiply by the square root of 10 over itself to the original equation.

I then got 40 inside of a square root symbol over 10. This is obviously wrong.<-- Why do you think so?

Factorize 40 and see if you can take some number outside the radicand sign. Remember

\(\displaystyle \sqrt{8} \,=\, 2\cdot\sqrt{2}\)


Start a new thread with new problem.
 
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