Q. 184.8( 4 Votes )

If a and b are the roots of x2 – 3x + p = 0 and c and d are the roots of x2 – 12x + q = 0, where a, b, c, d from a GP, prove that (q + p): (q – p) = 17: 15.

Answer :

Given data is,


x2 – 3x + p = 0 (1)


a and b are roots of (1)


So, (x + a)(x + b) = 0


x2 - (a + b)x + ab = 0


So, a + b = 3 and ab = p (2)


Given data is,


x2 – 12x + q = 0 (3)


c and d are roots of (1)


So, (x + c)(x + d) = 0


x2 - (c + d)x + cd = 0


So, c + d = 12 and cd = q (4)


a, b, c, d are in GP.(Given data)


Similarly A, AR, AR2, AR3 also forms a GP, with common ratio R.


From (2),


a + b = 3


A + AR = 3


(5)


From (4),


c + d = 12


AR2 + AR3 = 12


AR2 (1 + R) = 12 (6)


Substituting value of (1 + R) in (6).


R = 2


Now, substitute value of R in (5) to get value of A,


A = 1


Now, the GP required is A, AR, AR2, and AR3


1, 2, 4, 8…is the required GP.


So,


a = 1, b = 2, c = 4, d = 8


From (2) and (4),


ab = p and cd = q


So, p = 2, and q = 32.



So, (q + p): (q – p) = 17: 15.


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