this 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
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=?
this 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
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
take a eqn like dis
2x2-x-1=0
then...[5]
may b a stupid post[11]
i am also shocked... by the answer.. and they have asked it for 4 years.... :P
@abhirup... nice!! try that way...
yeah 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...
Let 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!!
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.
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
its 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 )
When the roots are real then it is not 0.
like...for the eqn 2x2-x-1=0 ...you can check this.
α+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