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Algebra

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21 Oct 2008 19:50:08 IST

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Solve for x, given that a,b,c>0

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Soumik

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22 Oct 2008 22:32:53 IST

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I'm not finding any solutions except x = 0...and I even have a proof that there aren't any either (it may be wrong)...but Bhatt sir, plz tell whether there are any?

Hari Shankar

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24 Oct 2008 14:15:38 IST

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hey guys, this aint so difficult. pls try

sutanoy dasgupta

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26 Oct 2008 19:28:32 IST

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simple inspection shows that x=0 is a solution.we prove that it is the only solution.

let f(x)= (a+bx)^1/2+(c+ax)^1/2+(b+cx)^1/2-(b-ax)^1/2-(a-cx)^1/2-(c-bx)^1/2.

differentiating with respect to x.

2f '(x)=b/(a+bx)^1/2 + a/(c+ax)^1/2 + c/(b+cx)^1/2 + a/(b-ax)^1/2 + c/(a-cx)^1/2 + b/(c-bx)^1/2.

as a,c,b >0, we conclude that f '(x)> 0. for all x.

as we are working in reals, all the terms a-bx,a-cx,c-bx,a+bx,... are positive, and we are considering the

positive roots of the terms as per convention.

also note that all the terms of f(x) can not be 0 at the same time.so at those points,f(x) will simply be equal to the nonzero terms, whose derivative is nonzero again.

now as f '(x) >0 , we conclude that the function f(x) is increasing strictly and can be 0 at only 1 point.

at x =0.so at x=0, the equality holds at x=0 which is the only solution.

Am I correct?

Hari Shankar

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26 Oct 2008 19:54:20 IST

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That is right.

But, what you can do after noting that x=0 is a solution, is to see that if x>0 (obviously within bounds) then $LHS>\sqrt a + \sqrt b + \sqrt c$ while $RHS<\sqrt a + \sqrt b + \sqrt c$

The reverse happens if x<0. Hence never the twain shall meet if $x \ne 0$

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