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 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 February 22nd, 2012, 04:46 AM #1 Newbie   Joined: Oct 2011 Posts: 3 Thanks: 0 How to separate x or solve for x w=a*b*c^3*x*?(pie^2 * (i-x^2)^2 + 16 *x^2)/ 4*d^2 *(1-x^2)^2 I have to find x ,i have the values of all other constants , I tried to separate it using partial fraction but I am stuck. February 22nd, 2012, 05:59 AM #2 Global Moderator   Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: How to separate x or solve for x I added some LaTex markup. If you would like help in solving problems, you should at least make it legible! See the link below. You can "Quote" this reply, then edit my math equation and use the preview button to make sure you got it right. February 22nd, 2012, 06:35 AM #3 Newbie   Joined: Oct 2011 Posts: 3 Thanks: 0 Re: How to separate x or solve for x [quote="arslan894"][quote="The Chaz"] the correct equation is like this . February 22nd, 2012, 07:19 AM   #4
Math Team

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Re: How to separate x or solve for x

Quote:
Originally Posted by arslan894
Quote:
Originally Posted by arslan894
Quote:
 Originally Posted by The Chaz the correct equation is like this .
Yikes! Why include a lot of variables to make the thing messier?
Let u = 4wd^2, v = abc^3 and p = (pi)^2

u(1 - x^2)^2 = vxSQRT[p(1 - x^2)^2 + 16x^2] ; square both sides:

u^2(1 - x^2)^4 = v^2 x^2[p(1 - x^2)^2 + 16x^2]

u^2(1 - x^2)^4 = p v^2 x^2(1 - x^2)^2 + 16v^2 x^4

I quit! February 22nd, 2012, 07:31 AM #5 Global Moderator   Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: How to separate x or solve for x Denis, don't make me give you a lecture about LaTex, too!  February 22nd, 2012, 01:23 PM #6 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 Re: How to separate x or solve for x Very drole, Sir Chaz! So we can get down to this (if I haven't goofed): u^2(1 - x^2)^4 = p v^2 x^2(1 - x^2)^2 + 16v^2 x^4 If it had turned out like: A(1 - x^2)^4 + B(1 - x^2)^2 + C = 0 then we'd have a quadratic, and solve by something like k = (1 - x^2)^2 : Ak^2 + Bk + C = 0 ....soooooo I wonder if there's a typo in the original problem. Tags separate, solve Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post sarah1994 Algebra 2 October 29th, 2013 12:16 PM The Chaz Calculus 1 August 5th, 2011 09:03 PM mbradar2 Calculus 1 December 2nd, 2010 09:03 AM gplush Algebra 1 August 25th, 2009 11:17 AM

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