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 December 16th, 2017, 07:18 PM #1 Newbie   Joined: Feb 2012 Posts: 19 Thanks: 0 Anyone can find the roots of this quadratic equation? $\displaystyle 2a^{2}+2b^{2}c^{2}+12bck +19k^{2}-2bc\cdot\sqrt{k\cdot\left(4bc+5k\right)}-7k\cdot\sqrt{k\cdot\left(4bc+5k\right)}+2a\cdot(-2bc-5k+\sqrt{k\cdot\left(4bc+5k\right)})=0$ Here a,b,c>0, and a>bc. I want to find the solution of k as a function of (a, b, c). Anyone can help? Thanks. Last edited by loveinla; December 16th, 2017 at 07:28 PM.
 December 16th, 2017, 07:28 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,935 Thanks: 1129 Math Focus: Elementary mathematics and beyond I don't think it's quadratic. Have you tried plugging it in to an online CAS (Computer Algebra System), such as Wolfram Alpha?
 December 17th, 2017, 07:30 AM #3 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 639 Thanks: 85 Can a quadratic equation have more than two variables?
 December 17th, 2017, 09:04 AM #4 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,414 Thanks: 1024 Yikes...one of ugliest equation I've seen! OK; here's a similar "less complicated" ugly one: k^2 + k + sqrt(ak) - sqrt(bk) - sqrt(ck) = 0 Go here to see solutions: http://www.wolframalpha.com/input/?i...k)+%3D+0+for+k Ready to quit?! Last edited by Denis; December 17th, 2017 at 09:09 AM.
December 17th, 2017, 01:19 PM   #5
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Quote:
 Originally Posted by EvanJ Can a quadratic equation have more than two variables?
Yes.

December 17th, 2017, 01:19 PM   #6
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Quote:
 Originally Posted by greg1313 I don't think it's quadratic. Have you tried plugging it in to an online CAS (Computer Algebra System), such as Wolfram Alpha?

 December 17th, 2017, 01:31 PM #7 Global Moderator   Joined: May 2007 Posts: 6,759 Thanks: 696 The square root expression is the same in all the terms containing it, so it can be factored out. Put the square root containg term on one side of the equation and the rest on the other side. Square both sides and you have a fourth degree polynomial in k.

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