Q. 18

If a and b are the roots of x2 – 3x + p = 0 and c, d are roots of x2 – 12x + q = 0, where a, b, c, d form a G.P. Prove that (q + p): (q – p) = 17:16.

Answer :

Given that a and b are roots of x2−3x + p=0


a + b = 3 and ab= p …(1)


[ If α and β are roots of the equation ax2 + bx + c=0 then α + β=−b/a and αβ=c/a.]


It is given that c and d are roots of x2−12x + q=0


c + d = 12 and cd=q …(2)


[ If α and β are roots of the equation ax2 + bx + c=0 then α + β=−b/a and αβ=c/a.]


Also given that a, b, c, d are in G.P.


Let a, b, c, d be the first four terms of a G.P.


So, a = a


b = ar


c = ar2


d = ar3


Now,


L.H.S






Now, From (1)


a + b=3


a + ar=3


a(1 + r)=3 ...(3)


From (2),


c + d=12


ar2 + ar3=12


ar2(1 + r)=12 ...(4)


Dividing equation (4) by (3), we get -


r2 = 4


r4 = 16


putting the value of r4 in L.H.S, we get -




Hence proved.

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