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January 8th, 2017, 05:49 AM   #1
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Simplify the monster

Ok so I woke up this morning to find this thing at the bottom of my bed. I've an exam tomorrow. This was from a previous paper. First bloody question on the exam. The poor critters.

I had to take a pic because the thing was to hard to write



Sorry I know it's a lil big. The picture I mean.

Now then so I started by getting rid of them square roots.

Ending up with just (c) on the numerator and b^2 on the bottom

I was thinking to myself - Ok better distribute out the top part see what we've got. But that just ends up a proper mess.

So I was thinking ok put them in brackets beside each other.
(3a^1/2 x b^-1 x c) (3a^1/2 x b^-1 x c)

Perhaps then I could use the powers on the bottom and just start subtracting away from the powers of the first term on the top...?

Or maybe I need to multiply out the term using a common denominator. But then that ends up a mess as well.

Can someone please push me in the general right direction. Thanks guys sorry I'm such a n00b.

Last edited by Kevineamon; January 8th, 2017 at 06:34 AM.
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January 8th, 2017, 06:55 AM   #2
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Had an idea. I was thinking using the power rule I could add up those powers in the two terms I've separated. Lets see hmmm

(3a^1/2 x b^-1 x c) (3a^1/2 x b^-1 x c)

3a x 2b^-2 x c^2

Hmmm is that right?
Can I just do that?
Would avoid the distribution
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January 8th, 2017, 07:02 AM   #3
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Got a headache as soon as I saw that !

Soooooo:
http://www.wolframalpha.com/input/?i...4)*c%5E(-1%2F3))

I entered:
x=(3*sqrt(a)*b^(-1)*sqrt(c^3))^2/(a^(-7)*sqrt(b^4)*c^(-1/3))
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Last edited by Denis; January 8th, 2017 at 07:07 AM.
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January 8th, 2017, 07:06 AM   #4
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$\Large \dfrac{\left(3a^{1/2}b^{-1}\sqrt{c^3}\right)^2}{a^{-7}\sqrt{b^4}c^{-1/3}}=$

$\Large \dfrac{\left(3 a^{1/2} b^{-1}c^{3/2}\right)^2}{a^{-7}b^2 c^{-1/3}}=$

$\Large \dfrac{9 a b^{-2} c^3}{a^{-7}b^2 c^{-1/3}}=$

$\Large 9a^8 b^{-4}c^{10/3} = $

$\Large \dfrac{9 a^8 c^{10/3}}{b^4}$
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Last edited by romsek; January 8th, 2017 at 07:25 AM.
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January 8th, 2017, 07:12 AM   #5
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There really is no way to avoid the work, but there really is not that much if you use PEMDAS.

$\dfrac{ \left ( 3a^{(1/2)}b^{-1}\sqrt{c^3} \right )^2}{a^{-7}\sqrt{b^4}c^{-1/3}} =$

$\dfrac{9a b^{-2} c^3}{a^{-7} b^2 c^{-1/3}} =$

$\dfrac{9a * a^7 * c^3 * \sqrt[3]{c}}{b^2 * b^2} =$

$what?$
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January 8th, 2017, 07:14 AM   #6
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Romsek, in denominator: c^(-1/3), not c^(1/3)
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January 8th, 2017, 07:38 AM   #7
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Thanks guys - interesting some of my ideas seemed to be correct, which shows my math brain is at least partially working.
K let me stare at this thing for another while.

Cheers guys you're the best
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January 8th, 2017, 08:01 AM   #8
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Very interesting the way that

$\displaystyle c^3/c^{-1/3}

became $

$\displaystyle c^{10/3} $

Never knew that very interesting - thanks again guys
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January 8th, 2017, 09:32 AM   #9
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Ya...tattoo this on your wrist:

x^a / x^b = x^(a-b) ; 3^7 / 3^4 = 3^(7-4) = 3^3
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January 8th, 2017, 10:09 AM   #10
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$\Large \dfrac{9 a^8 c^{10/3}}{b^4}$

Question:
that of course implies that b<>0;
will most teachers deduct a point or more if that's
not specified as part of the answer?
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