January 8th, 2017, 06:49 AM  #1 
Member Joined: Nov 2016 From: Ireland Posts: 54 Thanks: 2  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 07:34 AM. 
January 8th, 2017, 07:55 AM  #2 
Member Joined: Nov 2016 From: Ireland Posts: 54 Thanks: 2 
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 
January 8th, 2017, 08:02 AM  #3 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 8,136 Thanks: 552 
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)) Last edited by Denis; January 8th, 2017 at 08:07 AM. 
January 8th, 2017, 08:06 AM  #4 
Senior Member Joined: Sep 2015 From: CA Posts: 749 Thanks: 398 
$\Large \dfrac{\left(3a^{1/2}b^{1}\sqrt{c^3}\right)^2}{a^{7}\sqrt{b^4}c^{1/3}}=$ $\Large \dfrac{\left(3 a^{1/2} b^{1}c^{3/2}\right)^2}{a^{7}b^2 c^{1/3}}=$ $\Large \dfrac{9 a b^{2} c^3}{a^{7}b^2 c^{1/3}}=$ $\Large 9a^8 b^{4}c^{10/3} = $ $\Large \dfrac{9 a^8 c^{10/3}}{b^4}$ Last edited by romsek; January 8th, 2017 at 08:25 AM. 
January 8th, 2017, 08:12 AM  #5 
Senior Member Joined: May 2016 From: USA Posts: 466 Thanks: 198 
There really is no way to avoid the work, but there really is not that much if you use PEMDAS. $\dfrac{ \left ( 3a^{(1/2)}b^{1}\sqrt{c^3} \right )^2}{a^{7}\sqrt{b^4}c^{1/3}} =$ $\dfrac{9a b^{2} c^3}{a^{7} b^2 c^{1/3}} =$ $\dfrac{9a * a^7 * c^3 * \sqrt[3]{c}}{b^2 * b^2} =$ $what?$ 
January 8th, 2017, 08:14 AM  #6 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 8,136 Thanks: 552 
Romsek, in denominator: c^(1/3), not c^(1/3)

January 8th, 2017, 08:38 AM  #7 
Member Joined: Nov 2016 From: Ireland Posts: 54 Thanks: 2 
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 
January 8th, 2017, 09:01 AM  #8 
Member Joined: Nov 2016 From: Ireland Posts: 54 Thanks: 2 
Very interesting the way that $\displaystyle c^3/c^{1/3} became $ $\displaystyle c^{10/3} $ Never knew that very interesting  thanks again guys 
January 8th, 2017, 10:32 AM  #9 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 8,136 Thanks: 552 
Ya...tattoo this on your wrist: x^a / x^b = x^(ab) ; 3^7 / 3^4 = 3^(74) = 3^3 
January 8th, 2017, 11:09 AM  #10 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 8,136 Thanks: 552 
$\Large \dfrac{9 a^8 c^{10/3}}{b^4}$ 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? 

Tags 
monster, simplify 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Simplify  jiasyuen  Trigonometry  9  March 2nd, 2015 02:54 PM 
Simplify  jiasyuen  Algebra  1  December 28th, 2014 08:15 AM 
Simplify Cos  Shamieh  Algebra  2  September 12th, 2013 06:05 AM 
Simplify (?)  thinkfast  Algebra  2  January 27th, 2012 07:51 AM 
Simplify Cos  Shamieh  Calculus  2  January 1st, 1970 12:00 AM 