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December 16th, 2017, 08:18 PM   #1
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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 08:28 PM.
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December 16th, 2017, 08:28 PM   #2
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I don't think it's quadratic. Have you tried plugging it in to an online CAS (Computer Algebra System), such as Wolfram Alpha?
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December 17th, 2017, 08:30 AM   #3
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Can a quadratic equation have more than two variables?
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December 17th, 2017, 10:04 AM   #4
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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.
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December 17th, 2017, 02:19 PM   #5
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Quote:
Originally Posted by EvanJ View Post
Can a quadratic equation have more than two variables?
Yes.
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December 17th, 2017, 02:19 PM   #6
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Quote:
Originally Posted by greg1313 View Post
I don't think it's quadratic. Have you tried plugging it in to an online CAS (Computer Algebra System), such as Wolfram Alpha?
It's certainly not a quadratic.
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December 17th, 2017, 02:31 PM   #7
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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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