sqrt (1/12) - sqrt (1/27)

- anonymous

sqrt (1/12) - sqrt (1/27)

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- anonymous

When you have the sqrt of a fraction, it is the sqrt of the numerator divided by the sqrt of the denominator. Can you convert both expressions to a fraction with denominator 6?

- anonymous

Yes I can. The first thing you want to do is get the 1's out of the squareroot.
\[\sqrt{a/b}=\sqrt{a}/\sqrt{b}\]
So \[\sqrt{1/b}=\sqrt{1}/\sqrt{b}=1/\sqrt{b}\]

- anonymous

o okay

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## More answers

- anonymous

Does that make sense or did I use too many symbols and make it too long winded?

- anonymous

no it makes sense thank u!

- anonymous

The next thing to do is to simplify \(\sqrt{12},\sqrt{27}\)

- anonymous

do you know how to do that?

- anonymous

find a common denominator?

- anonymous

Basically, what he's saying is that 1/sqrt(a) = sqrt(a) / a. This is because you can multiply by sqrt(a)/sqrt(a) and the root on the bottom cancels out, while the 1 on the top becomes sqrt(a). You get sqrt(12)/12 - sqrt(27)/27, and you know how to simplify from there.

- anonymous

That's probably a simpler way to look at it, I was going to simplify the denominator and then move it to the top.

- anonymous

But regardless, you need to simplify the radical.

- anonymous

but they dont reely have a common denominator do they?

- anonymous

would it be 3 * 4 and 3 * 9 for it?

- anonymous

In order to simplify a radical like \(\sqrt{12}\), you need to find a square that divides it. In this case, 4 divides 12, so we write
\[\sqrt{12}=\sqrt{4}\sqrt{3}=2\sqrt{3}\]

- anonymous

Right, exactly.

- anonymous

Would you prefer to move everything to the top and then deal with it or deal with it first, then move it to the top?

- anonymous

either way is easiest

- anonymous

i mean whichever way is easiest

- anonymous

Let's do it the way quantummodulus suggested. Did you understand what they said?

- anonymous

yes

- anonymous

Ok. So what have you gotten to so far?

- anonymous

(sqrt 12/2 sqrt 3) - (sqrt 27/3 sqrt 3)

- anonymous

Ok, not quite. I think we got you jumbled up a bit. Let's take a step back. First, we're going to do what is called rationalizing the denominator. That is where we multiply the top and the bottom by the denominator. This will get rid of the square root. Do you know how to do this?

- anonymous

yes

- anonymous

Ok, so what do you get after rationalizing the denominator?

- anonymous

i think i square rooted something that didnt need to be sqaure rooted

- anonymous

so now i have sq rt 12/12 - sqrt 27/27 the denominators not being sqrt

- anonymous

Yup, perfect. Now, simplify the radical, like you did before.

- anonymous

on top?

- anonymous

Yup.

- anonymous

okay i got 2 sqrt 3/12 - 3 sqrt 3/27

- anonymous

Right, now cancel and add fractions.

- anonymous

sqrt 3/6 - sqrt3/9

- anonymous

Yup, now common denominators and add.

- anonymous

3/18 - sqrt 6/18

- anonymous

\(3*\sqrt{3}\) doesn't equal 3, same for the other fraction

- anonymous

3 sqrt 3/18 - 2sqrt 3/9

- anonymous

/18

- anonymous

now just subtract

- anonymous

sqrt 3/18 is the answer then?

- anonymous

Yup. So the process is: rationalize the denominator, simplify the radical, simplify fractions

- anonymous

Sorry about all of the confusion at the beginning.

- anonymous

oh no that was my fault thank u for all your help thank u!

- anonymous

You're welcome. Good luck on any other problems.

- anonymous

\[\sqrt{a}-\sqrt{b} = \sqrt{a-2\sqrt{a*b}+b}\]
Plug in 1/12 for a and 1/27 for b and simplify. The result is:
\[{\sqrt{1\over 108}} = {1 \over 6\sqrt{3}}\]
The formula is obtained by squaring the quantity radical a - radical b, ie: the left side of the formula equation.
The first and last terms of the result is a and b respectively. The mid term is negative two times the product of radical a and radical b. Radical a times radical b is equal to the square root of the product of a and b.
The final operation is to take the square root of every thing.

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