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#1 Posted 0:50pm 11-02-09 Quad... answer only...Don't want soln or any hint just the answer.. If the roots of the equation ax[p]2[/p]+cx+c=0 be in ratio p:q then.... [sqrt]p/q[/sqrt]+[sqrt]q/p[/sqrt]+[sqrt]c/a[/sqrt]=?
http://priyamsays.blogspot.com |
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#2 Posted 0:58pm 11-02-09 Re: Quad... answer only...2√2[i][b]i [/b][/i]???
http://abhirupsarkar.webs.com/ Edited on 0:09pm 12-02-09 |
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#3 Posted 0:59pm 11-02-09 Re: Quad... answer only...lol...
http://priyamsays.blogspot.com |
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#4 Posted 1:01pm 11-02-09 Re: Quad... answer only...this is from B.U. 77,64 R.U. 69,75 and they asked to prove it is equal to [b]zero[/b].. :D thats why i posted this question... [4] In hall and Knight too. same question... and in few books they proved it to be zero by two methods.. lol
http://priyamsays.blogspot.com |
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#5 Posted 1:06pm 11-02-09 Re: Quad... answer only...howz dat possible???????????????? dat means all p/q , q/p , c/a hav 2 be individually zero as their sq roots cant b -ve..... which is impossible as if p/q =0, means p=0; then how can q/p=0 it shud b ∞.... I M SHOCKD by the answer |
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#6 Posted 1:06pm 11-02-09 Re: Quad... answer only...take a eqn like dis 2x[p]2[/p]-x-1=0 then...[5] may b a stupid post[11]
http://abhirupsarkar.webs.com/ |
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#7 Posted 1:10pm 11-02-09 Re: Quad... answer only...Priyam ..post the soln plzzzz....
http://abhirupsarkar.webs.com/ |
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#8 Posted 1:11pm 11-02-09 Re: Quad... answer only...i am also shocked... by the answer.. and they have asked it for 4 years.... :P @abhirup... nice!! try that way...
http://priyamsays.blogspot.com |
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#9 Posted 1:15pm 11-02-09 Re: Quad... answer only...yeah tapan i also thought the same... each must be zero... so p=q=c=0 then eqn is ax[p]2[/p]=0 and about soln given in book dekhte hain.. kya kiya hai...
http://priyamsays.blogspot.com |
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#10 Posted 1:45pm 11-02-09 Re: Quad... answer only...are you sure you have written down the question properly? |
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#11 Posted 1:54pm 11-02-09 SolnLet the roots be kp and kq (k is an arbitrary constant); Then, √(p/q)+√(q/p)+√(c/a)= (p+q)/√(pq) + √(c/a)= (kp+kq)/√(kp.kq) + √(c/a)= Sum of rts/√(Prod of rts) + √(c/a)= (-c/a)/√(c/a) + √(c/a)= -√(c/a) + √(c/a) = 0!!
Edited on 2:30pm 11-02-09 |
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#12 Posted 1:55pm 11-02-09 Re: Quad... answer only...priyam verify hall and knight once more |
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#13 Posted 1:56pm 11-02-09 Cont.U gus hav ignored the fact that the roots can be imaginary, in which case, the root of the ratio of the roots can be -ve. |
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#14 Posted 2:16pm 11-02-09 Re: Quad... answer only...if you see correctly, then the given sum will be √(p/q)+√(q/p)+√(c/a)= (|p|+|q|)/√(pq)| + √(c/a)= (k|p+k|q|)/√(kp.kq) + √(c/a) whcih will not be the thing given unless we know that the roots are not real! Does this make more sense?!? (I am not very sure bcos i dont want to take risk of trying to defy a classic like Hall and Knight.) Opinions invited :) Great that you noticed this abhishek
-when spring comes, it melts the snow one flake at a time… |
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#15 Posted 4:56pm 11-02-09 Re: Quad... answer only...its not 0 the fact is p/q is ω or ω[p]2[/p] and proceed ( the fallacy is wen u take √a√a = a wen its actually -a ) |
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#16 Posted 0:13pm 12-02-09 Re: Quad... answer only...When the roots are real then it is not 0. like...for the eqn 2x[p]2[/p]-x-1=0 ...you can check this.
http://abhirupsarkar.webs.com/ |
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#17 Posted 0:58pm 14-02-09 Re: Quad... answer only...α+β=-c/a αβ=c/a α+β=-αβ =>α+β=[b]-[sqrt]αβ[/sqrt]*[sqrt]αβ[/sqrt][/b] =>[sqrt]α/β[/sqrt]+[sqrt]β/α[/sqrt]=-[sqrt]αβ[/sqrt] => √p/q+√q/p+√c/a=0!!! problem is there in 2nd line when we consider -αβ= -[sqrt]αβ[/sqrt]*[sqrt]αβ[/sqrt] as here we r only considering both+ve square roots of αβ
http://abhirupsarkar.webs.com/ Edited on 0:59pm 14-02-09 |
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