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 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 $w=a \cdot b \cdot c^3 \cdot x \cdot \sqrt{(\pi^2 \cdot (i-x^2)^2 + 16 \cdot x^2)/ 4 \cdot d^2 \cdot (1-x^2)^2}$ 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"]$w \cdot 4 \cdot d^2 \cdot (1-x^2)^2=a \cdot b \cdot c^3 \cdot x \cdot \sqrt{(\pi^2 \cdot (1-x^2)^2 + 16 \cdot x^2)}$ 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 $w \cdot 4 \cdot d^2 \cdot (1-x^2)^2=a \cdot b \cdot c^3 \cdot x \cdot \sqrt{(\pi^2 \cdot (1-x^2)^2 + 16 \cdot x^2)}$ 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: 14,594 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.

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