36What wrong do u find??
if, a1/b1 = a2/b2 then,
a1/b1 = a2/b2 = (a1 + a2)/(b1 + b2) = √(a1.a2)/(b1.b2)
verification
2/4 = 4/8
then, 2/4 = 4/8 = 2+4/4+8 = √(2x4)/(8x4)
i think u meant to comment on this step.....!!
24 Answers
36
1I am using Delta=D ,alpha=a ,beta=b
Dear Rahulmishra
Why are taking a+b=0,as you are showing B=0?
We are getting a-b=0 not a+b=0.
So either D=0 or c=0.
It is not like this that D is always zero.It is not like this that c is always equal to zero.It is like this that D may be zero.It is like this that c may be zero.
Now consider,cD=0.
In this case one has to be zero.Suppose that we take c to be equal to zero then (a) is correct.
Now consider the case of option (c)
bD=0.In this case D need not to be equal to zero.And we do not know about the value of b.So we cannot say anything about it.
So (b) need not to be correct.So the answer is only option (a).
36okay....
αβ(α2 + β2) = 2 α2β2
=> αβ(α - β)2 = 0
so either,
αβ = 0 or α = β
=> α = 0 and β = 0
Now, αβ = c/a
=> c = 0
also, α + β = -b/a
=> b = 0
But what does that have to take with
a∆ = 0 and b∆ = 0 .... tell me...!!
1Here a=alpha ,b=beta
from your equation
2a2b2=ab(a2+b2)
Let us do not cancel it
ab(a2+b2-2ab)=0
=>ab(a-b)2=0
It means
either ab=0 or (a-b)2=0
ab=0>c=0
(a-b)2=0>delta=0
These are the results that we get.We do not get c=0 when we cancel ab.ok?
36kk...
but thats wonderful of you..
we must always accept the truth and decline the false... coz.. that leads to success...!!
Thumbs up for u.. from my side...!!!
1nonono.I always accept the truth.I were not familiar with this result.
71Pata nai kyo mujhe isme kuch problem lag rahi hai :( :(
36Since,
α + β, α2 + β2, α3 + β3 are in G.P
=> α2 + β2 / α + β = α3 + β3/ α2 + β2 = √(α2 + β2)(α3 + β3)/(α + β)(α2 + β2)
= √(α3 + β3)/(α + β)
= √(α2 + β2 - αβ)
=> α2 + β2/α + β = √(α2 + β2 - αβ)
on squaring both the sides we get,
(α2 + β2)2 = (α + β)2 . (α2 + β2 - αβ)
=> (α2 + β2)2 = (α2 + β2 + 2 αβ) . (α2 + β2 - αβ)
=> (α2 + β2)2 = (α2 + β2)2 - αβ (α2 + β2) + 2 αβ (α2 + β2) - 2 α2β2
=> αβ(α2 + β2) = 2 α2β2
=> α2 + β2 = 2 αβ
Again,
α2 + β2 = (α + β)2 - 2αβ
=>4αβ =(α + β)2
=> 4c/a = b2/a2
=> b2 = 4ac
=> b2 - 4ac = 0
=> ∆ = 0
Thus, (a) and (c) both must be correct....!!
1ok see i will explain it to u in detail now....u wrote the G.P. condition above[y2=xz one right??]
now i m using these symbols a=alpha b=beta,determinant=D
on simplifying ur G.P. condition we get this
ab(a-b)2=0[agree??]
so either
1.ab=0 OR
2.a-b=0
1. means D=0
2.means c=0[ note a≠0...else it won't be a quadratic]
so summing up either c=0 or D=0
not i highlighted word OR every where...since all we can conclude is either c=0 OR D=0
we can't say that D=0 always....all we can say either c=0 OR D=0 so it means c*D=0 hence only (a) is correct...since we can'rt say D=0 always so (b) is not right choice
1Dear Rahulmishra
I have proved it on the top.Read the calculation below the question very carefully.I couldn't do detailed working.But you will find that it has been proved to be zero,.
1Sorry for that.But if delta=0 then why not b.delta=0 and why just c.delta=0
1Dear Nishant Sir
From your first equation ab=0=>a=0 or b=0
Then why not b.Delta=0.That means why not option (c) is also correct.This is my question.
622a^2b^2=ab(a^2+b^2)
Now you have cancelled ab without any thought...
ab = c/a
here ab is being used as alpha and beta respectively..