Latest Wipro Aptitude Question SOLUTION: If the sum of the roots of the equation ax²+bx+c=0 is equal to the sum of the squares of their reciprocals, then, a/c, b/a, c/b are in (a) AP (b) GP (c) HP (d) None of the

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If the sum of the roots of the equation ax²+bx+c=0 is equal to the sum of the squares of their reciprocals,
then, a/c, b/a, c/b are in

(a) AP
(b) GP
(c) HP
(d) None of the these

Asked In Wipro jaya mishra (11 years ago)

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Aman  (11 years ago)

a/c, b/a, c/b are in HP... option C

Is this solution Helpfull? Yes (10) | No (2)  

Ganga Prasad Bhusal  (11 years ago)

Let roots be P and q
then p+q=(1/p^2)+(1/q^2)=(p^2+q^2)/(P^2*q^2)
p+q = ((p+q)^2 - 2*p*q)/(p*q)^2-----------------(1)
we know that sum of roots p+q = -b/a
and product of roots p*q = c/a
putting these values in eqn (1), we get-
-b/a = ((-b/a)^2 - 2*c/a)/(c/a)^2
(-b/a)*(c^2/a^2)= (b^2/a^2)- 2*c/a
multiplying both sides with a^3,we get-
a*b^2 - 2*c*a^2 = -b*c^2
dividing both sides again by abc, we get--
2*a/b = b/c + c/a
b/(2*a) = a*c/(ab + c^2)
b/a = (2*(a/c)*(c/b))/(a/c + c/b)-----------------(2)
we know that three no. k,l,m are in h.p. if l = (2*k*m)/(k+m)
So from eqn (2), we can say that a/c,b/a,c/b are in h.p.
so ANSWER = option (c)

Is this solution Helpfull? Yes (7) | No (2)  

satyaprakash vengala  (11 years ago)

i get answer b^2=a^2.
but i dont knw they r in a.p r g.p r h.p?
anyone can explain?

Is this solution Helpfull? Yes (1) | No (5)  

lakshmi  (11 years ago)

ap option a

Is this solution Helpfull? Yes | No (10)  

Niraj Kumar  (7 years ago)

Answer is HP.

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