Quadratic....

The following is a system of quadratic equation with a,b,c being real and a≠0.....

ax12+bx1+c=x2

ax22+bx2+c=x3

ax32+bx3+c=x4

ax42+bx4+c=x5
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.
.
.
.
.
.
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axn-12+bxn-1+c=xn

axn2+bxn+c=x1

Answer the following questions....

Q1) The no of real solutions if (b-1)2<4ac ...
Q2) The no of real solutions if (b-1)2>4ac ...
Q3) The no of real solutions if (b-1)2=4ac ...

16 Answers

62
Lokesh Verma ·

this is an awesome question...

If i remember correctly, it was solved by prophet sir, sometime back.. let me dig the post out!

341
Hari Shankar ·

no i didnt. but i remember the qn from my student days. was in TMH if my memory serves me right

62
Lokesh Verma ·

it is from Arihant..

SOmeone asked me recently.. I had a hard tiem to explain why this is not obviously 1)
and it took me a lot of trouble to make him understand that the question was not obvious!!

3
msp ·

1) zero solns is dat rite.

62
Lokesh Verma ·

yes.. why?

3
msp ·

sir adding the eqns i substitute y=Σxi

we will have a quadratic eqn the quadratic eqn in y have no roots

62
Lokesh Verma ·

no sankara.. this is not the right way..

see the expression again..

11
Aditya Balasubramanyam ·

The main hitch in solving this question is identifying that (b-1)2-4ac is the discriminant....

62
Lokesh Verma ·

@adiya .. yes...

But that alone will not help!

what I mean that is what gave me a tough time to explain to one of the students that (b-1)2-4ac does not mean that none of these equations have a solution!

Because what he was arguing is that none of these equations have a solution.. hence the whole set of these equations cannot have a solution!

11
Aditya Balasubramanyam ·

yes definitely... But any hints on solving this toughie ?

62
Lokesh Verma ·

Think what happens to the sum...

11
Aditya Balasubramanyam ·

It did'nt Lead To Anything

11
Devil ·

Experts pls wait for 1 more day.....

62
Lokesh Verma ·

aditya... express teh sum of all the expressions above as a sum of quadratics

11
Devil ·

axi2+bxi+c=xi+1

If (b-1)2-4ac<0....then subtract xi from both sides....

Now add all the equations after this operation.
L.H.S is not 0...(meaning it's either >0 or <0, depending on a)....while R.H.S=0
Thus no solutions.

62
Lokesh Verma ·

Yes soumik :)

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