anonymous
  • anonymous
sqrt (1/12) - sqrt (1/27)
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • 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
  • 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
  • anonymous
o okay

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anonymous
  • anonymous
Does that make sense or did I use too many symbols and make it too long winded?
anonymous
  • anonymous
no it makes sense thank u!
anonymous
  • anonymous
The next thing to do is to simplify \(\sqrt{12},\sqrt{27}\)
anonymous
  • anonymous
do you know how to do that?
anonymous
  • anonymous
find a common denominator?
anonymous
  • 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
  • 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
  • anonymous
But regardless, you need to simplify the radical.
anonymous
  • anonymous
but they dont reely have a common denominator do they?
anonymous
  • anonymous
would it be 3 * 4 and 3 * 9 for it?
anonymous
  • 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
  • anonymous
Right, exactly.
anonymous
  • 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
  • anonymous
either way is easiest
anonymous
  • anonymous
i mean whichever way is easiest
anonymous
  • anonymous
Let's do it the way quantummodulus suggested. Did you understand what they said?
anonymous
  • anonymous
yes
anonymous
  • anonymous
Ok. So what have you gotten to so far?
anonymous
  • anonymous
(sqrt 12/2 sqrt 3) - (sqrt 27/3 sqrt 3)
anonymous
  • 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
  • anonymous
yes
anonymous
  • anonymous
Ok, so what do you get after rationalizing the denominator?
anonymous
  • anonymous
i think i square rooted something that didnt need to be sqaure rooted
anonymous
  • anonymous
so now i have sq rt 12/12 - sqrt 27/27 the denominators not being sqrt
anonymous
  • anonymous
Yup, perfect. Now, simplify the radical, like you did before.
anonymous
  • anonymous
on top?
anonymous
  • anonymous
Yup.
anonymous
  • anonymous
okay i got 2 sqrt 3/12 - 3 sqrt 3/27
anonymous
  • anonymous
Right, now cancel and add fractions.
anonymous
  • anonymous
sqrt 3/6 - sqrt3/9
anonymous
  • anonymous
Yup, now common denominators and add.
anonymous
  • anonymous
3/18 - sqrt 6/18
anonymous
  • anonymous
\(3*\sqrt{3}\) doesn't equal 3, same for the other fraction
anonymous
  • anonymous
3 sqrt 3/18 - 2sqrt 3/9
anonymous
  • anonymous
/18
anonymous
  • anonymous
now just subtract
anonymous
  • anonymous
sqrt 3/18 is the answer then?
anonymous
  • anonymous
Yup. So the process is: rationalize the denominator, simplify the radical, simplify fractions
anonymous
  • anonymous
Sorry about all of the confusion at the beginning.
anonymous
  • anonymous
oh no that was my fault thank u for all your help thank u!
anonymous
  • anonymous
You're welcome. Good luck on any other problems.
anonymous
  • 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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