13take a eqn like dis
2x2-x-1=0
then...[5]
may b a stupid post[11]
Don't want soln or any hint just the answer..
If the roots of the equation ax2+cx+c=0 be in ratio p:q then....
√p/q+√q/p+√c/a=?
-
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16 Answers
33this is from B.U. 77,64 R.U. 69,75 and they asked to prove it is equal to zero.. :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
21howz 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
1333i am also shocked... by the answer.. and they have asked it for 4 years.... :P
@abhirup... nice!! try that way...
33yeah tapan i also thought the same...
each must be zero...
so p=q=c=0
then eqn is ax2=0
and about soln given in book dekhte hain.. kya kiya hai...
1Let 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!!
1U 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.
62if 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
9its not 0
the fact is p/q is ω or ω2 and proceed
( the fallacy is wen u take √a√a = a wen its actually -a )
13When the roots are real then it is not 0.
like...for the eqn 2x2-x-1=0 ...you can check this.
13α+β=-c/a
αβ=c/a
α+β=-αβ
=>α+β=-√αβ*√αβ
=>√α/β+√β/α=-√αβ
=> √p/q+√q/p+√c/a=0!!!
problem is there in 2nd line when we consider -αβ= -√αβ*√αβ
as here we r only considering both+ve square roots of αβ