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It is given that a and b are the roots of x2 – 3x + p = 0
∴ a + b = 3 and ab = p … (1)
Also, c and d are the roots of
∴c + d = 12 and cd = q … (2)
It is given that a, b, c, d are in G.P.
Let a = x, b = xr, c = xr2, d = xr3
From (1) and (2), we obtain
x + xr = 3
⇒ x (1 + r) = 3
xr2 + xr3 =12
⇒ xr2 (1 + r) = 12
On dividing, we obtain
Case I:
When r = 2 and x =1,
ab = x2r = 2
cd = x2r5 = 32
Case II:
When r = –2, x = –3,
ab = x2r = –18
cd = x2r5 = – 288
Thus, in both the cases, we obtain (q + p): (q – p) = 17:15
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