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Square Roots Question

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Square Roots Question [#permalink] New post 24 Aug 2010, 09:36
Hello all,

Working through the MGMAT Equations, Inequalities, and VICs book and encounter a roots answer in chapter 4 that totally stumps me. The specific question is #9 on page 63.

You have to pick numbers for the question and end up with a fraction as follows: √1/2, (whole fraction under radical) which then becomes 1/√2 = √2/2 (only numerator under radical) -> wait what? what's the general property that I need to know here? Any square root with a numerator of 1 can just be rewritten by moving the radical to the denominator? And on top of that, how do you know to then make it √2/2 (only numerator under radical) ?? I guess I just havent seen this manipulation before and it just strikes me as hard to think through. It's not easy to see how they are equal.

On top of that, a further twist regarding the manipulation:

As you are working through the math you get to 2√3/√2 which is then manipulated as such: 2√3(√2)/2 ->huh? How did you move the sqrt of 2 from the bottom to the top (not only moving, but retaining it in the bottom as well)??

Any help would be greatly appreciated
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Re: Square Roots Question [#permalink] New post 24 Aug 2010, 13:17
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Hi HustleHarder,

One property of fraction is that the fraction remains same if you multiply or divide the numerator and denominator with the same value.

In √1/2, (whole fraction under radical) we can write it as √1/√2 ( another property of square rt. √(X/Y) = √X/√Y

now we know that √1 = 1 so the complete fraction turns out to 1/√2.
if we multiply both numerator and denominator by √2, we get √2/(√2 * √2)
here we have to apply another property of sq rt. √X*√Y = √(X*Y)

so the fraction becomes √2/√(2*2) = √2/√4 = √2/2 (since √4 = 2).

regarding, 2√3/√2 here we are multiplying both the numerator and denominator by the same value √2 so we get 2√3√2/√2√2 = 2√3√2 / √(2*2) = 2√3√2/ √4 = 2√3√2/2 = √6

-Please let me know if you need more description in any of these [:)]
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Re: Square Roots Question [#permalink] New post 24 Aug 2010, 14:27
Dude. Do you know you are the man? Thanks!!! Makes total sense now. Had the eureka moment. I may have some more soon...
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Re: Square Roots Question [#permalink] New post 24 Aug 2010, 15:04
No issues keep them posting ;)
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Re: Square Roots Question [#permalink] New post 24 Aug 2010, 19:56
Got another one here, this time compound functions where you have to find a value of x for which f(g(x)) = g(f(x)) given that f(x) = x^3 + 1 and g(x) = 2x. I set it up as follows:

f(2x) = g(x^3+1) now this is where the MGMAT book does another quantum leap in algebraic manipulation that leaves me dumbfounded. It takes the expression above and turns it to:

(2x)^3+1 = 2(x^3+1)

wuh wuh whaaa? They do not explain what they are doing to manipulate the equation. Help would be appreciated again : )
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Re: Square Roots Question [#permalink] New post 02 Sep 2010, 16:17
Hi HustleHarder,

Sorry for the long delay in posting a reply.

we know that f(x) = x^3 + 1 and g(x) = 2x.

that means for any value of x f(x) is (value of x)^3 +1 and g(x) is 2 times value of x.

so f(2x) becomes (2x)^3 + 1 and g(x^3+1) = 2 times (x^3 + 1)

please let me know if it helps.
Re: Square Roots Question   [#permalink] 02 Sep 2010, 16:17
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