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December 16th, 2017, 08:18 PM  #1 
Newbie Joined: Feb 2012 Posts: 19 Thanks: 0  Anyone can find the roots of this quadratic equation?
$\displaystyle 2a^{2}+2b^{2}c^{2}+12bck +19k^{2}2bc\cdot\sqrt{k\cdot\left(4bc+5k\right)}7k\cdot\sqrt{k\cdot\left(4bc+5k\right)}+2a\cdot(2bc5k+\sqrt{k\cdot\left(4bc+5k\right)})=0$ 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 08:28 PM. 
December 16th, 2017, 08:28 PM  #2 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,692 Thanks: 976 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, 08:30 AM  #3 
Senior Member Joined: Oct 2013 From: New York, USA Posts: 580 Thanks: 80 
Can a quadratic equation have more than two variables?

December 17th, 2017, 10:04 AM  #4 
Math Team Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 11,346 Thanks: 728 
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 10:09 AM. 
December 17th, 2017, 02:19 PM  #5 
Senior Member Joined: May 2016 From: USA Posts: 881 Thanks: 353  
December 17th, 2017, 02:19 PM  #6  
Senior Member Joined: May 2016 From: USA Posts: 881 Thanks: 353  Quote:
 
December 17th, 2017, 02:31 PM  #7 
Global Moderator Joined: May 2007 Posts: 6,416 Thanks: 557 
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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