March 5th, 2017, 08:45 PM  #1 
Newbie Joined: Feb 2017 From: USA Posts: 28 Thanks: 1  Simplifying Nested Fractions
Can anyone direct me where I went wrong with this example: $\displaystyle \frac{\sqrt{4x^{2}}\frac{1}{2\sqrt{4x^{2}}}}{4x^{2}}$ My first inkling was to simplify the top section by multiplying by the reciprocal of the top right fraction  by $\displaystyle {2\sqrt{4x^{2}}}$ to give like denominators. So left side should look like this: $\displaystyle {2\sqrt{4x^{2}}}*{\sqrt{4x^{2}}}$ Which should resolve to 2*$\displaystyle 4x^{2}$ The right side of the top of the original fraction should now resolve to 1 after we multiply with the reciprocal. We now have : $\displaystyle \frac{2*(4x^{2})1}{4x^{2}}$ I see the opportunity to cancel out like terms, removing the $\displaystyle 4x^{2}$ term from both sides of the fraction. That leaves $\displaystyle \frac{2*11}{1}$ Or simply 1. Too bad after all that work and 'logic' the answer is wrong and I really don't understand what I did wrong. I am trying to shore up my deficiencies in algebra so I can return to school. Calculus I was long ago for me, I took a B, but I am sure that I need to retake it at this point. Apparently algebra too!! Any assistance is greatly appreciated. No answers please, but directions on what went wrong and perhaps why. Thank you. 
March 6th, 2017, 05:33 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,291 Thanks: 1683 
You have to divide (or multiply) the numerator and denominator by the same thing. If the numerator (or denominator) contains more than one term, your division (or multiplication) must be applied to each term in the numerator (or denominator). You didn't apply the above rule consistently. 
March 6th, 2017, 06:07 AM  #3 
Newbie Joined: Feb 2017 From: USA Posts: 28 Thanks: 1 
!!! Did not cross multiply to make like terms for my top side of the equation! Thank you for the guidance, let's see what I make of it now: $\displaystyle \frac{\sqrt{4x^{2}}\frac{1}{2\sqrt{4x^{2}}}}{4x^{2}}$ Cross multiply only the top side: $\displaystyle \frac{\sqrt{4x^{2}}}{1}*\frac{2\sqrt{4x^{2}}}{2\sqrt{4x^{2}}}\frac{1}{2\sqrt{4x^{2}}}*\frac{1}{1}$ We are only cross multiplying the denominators to make equivalent denominators so that we can simplify the top fraction. From this I get: $\displaystyle \frac{\frac{\sqrt{4x^{2}}*2\sqrt{4x^{2}}1}{2\sqrt{4x^{2}}}}{2\sqrt{4x^{2}}}$ We can take out our like terms and get: $\displaystyle \frac{\sqrt{4x^{2}}1}{2\sqrt{4x^{2}}}$ Have I got it now? Last edited by skipjack; March 6th, 2017 at 06:27 AM. 
March 6th, 2017, 06:22 AM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,904 Thanks: 883 
You may find it easier this way: let u = 4  x^2 expression becomes: [sqrt(u)  1/(2sqrt(u))] / u simplify: you should get (2u  1) / (2u sqrt(u)) Substitute back in. Makes the step by step process easier... Note: you can assign a value to x (like x=1) and then evaluate the original expression and your results....get my drift? Last edited by Denis; March 6th, 2017 at 06:34 AM. Reason: added note 
March 6th, 2017, 06:35 AM  #5 
Global Moderator Joined: Dec 2006 Posts: 19,291 Thanks: 1683 
As chopnhack requested that we don't supply answers, I will limit my advice to saying that "take out our like terms" is a vague description that shouldn't be used as an alternative to giving the working in full (which would enable identification of any mistakes).

March 6th, 2017, 06:48 AM  #6 
Senior Member Joined: May 2016 From: USA Posts: 1,084 Thanks: 446 
Taking Denis's approach even farther $Let\ u = \sqrt{4  x^2} \implies \dfrac{\sqrt{4  x^2}  \dfrac{1}{2\sqrt{4  x^2}}}{4  x^2} = \dfrac{u  \dfrac{1}{2u}}{u^2}.$ 
March 6th, 2017, 06:54 AM  #7 
Newbie Joined: Feb 2017 From: USA Posts: 28 Thanks: 1 
Sorry for the vagueness. What I meant by remove the like terms is to divide out the $\displaystyle 2\sqrt{4x^{2}}$ from the third equation in my last post. It appeared in both the numerator and denominator of the top part of the nested fraction. I will also explore the substitution methods you fine gents are posting in a moment. I am still curious to work this out and see what I am doing wrong. Sometimes these complicated fractions have a way of getting me spun around and I need to nail this skill down for deriving future equations. Thanks! Last edited by chopnhack; March 6th, 2017 at 07:14 AM. 
March 6th, 2017, 07:13 AM  #8 
Newbie Joined: Feb 2017 From: USA Posts: 28 Thanks: 1 
On the u front: $\displaystyle \frac{\sqrt{4  x^2}  \frac{1}{2\sqrt{4  x^2}}}{4  x^2} = \frac{u  \frac{1}{2u}}{u^2}$ so this can further be simplified by creating the same denominator for the top half of the nested fraction: $\displaystyle \frac{u  \frac{1}{2u}}{u^2}$ becomes: $\displaystyle \frac{\frac{2u*u}{2u}\frac{1}{2u}}{u^{2}}$ becomes: $\displaystyle \frac{\frac{2u*u1}{2u}}{u^{2}}$ using the property of (a/b)/c is equivalent to a/bc gives: $\displaystyle \frac{2u*u1}{2u*u^{2}}$ from here I can see two paths to further resolve the equation, but one makes the equation zero, which I will omit since it doesn't make sense: $\displaystyle \frac{2u^{2}1}{2u^{3}}$ I see the ability to reduce down our u's: $\displaystyle \frac{2(1)1}{2u}$ or $\displaystyle \frac{1}{2u}$ = $\displaystyle \frac{1}{2\sqrt{4  x^2}}$ Unfortunately I know that this is not the correct answer. Please assist me in finding the flaw in the process, thanks! Last edited by chopnhack; March 6th, 2017 at 07:15 AM. 
March 6th, 2017, 08:51 AM  #9 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,904 Thanks: 883  
March 6th, 2017, 09:02 AM  #10 
Global Moderator Joined: Dec 2006 Posts: 19,291 Thanks: 1683 
The wording "reduce down our u's" is just as vague as "take out our like terms".


Tags 
fractions, nested, simplifying 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Simplifying fractions  mathmere  Algebra  3  October 12th, 2016 08:23 AM 
Simplifying fractions  ophion  Elementary Math  6  March 20th, 2016 05:17 PM 
simplifying with fractions  markcr  Algebra  6  June 3rd, 2014 01:22 PM 
Simplifying Fractions  manjit  Elementary Math  0  November 26th, 2010 02:07 PM 
Simplifying fractions with variables in fractions  nova3421  Algebra  1  September 9th, 2009 10:03 AM 