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Question
If a and b are the roots of are 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 : 15.
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Solution
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 x2 – 12x + q = 0
∴ 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
`(x^2 (1 + r))/(x (1 + r)) = (12)/(3)`
= r2 = 4
= r = ±2
When r = 2, `x = 3/(1 + 2) = 3/2 = 1`
When r = -2, `x = 3/(1 - 2) = 3/(-1) = -3`
Case I:
When r = 2 and x = 1
ab = x2 r = 2
cd = x2 r5 = 32
∴ `(q + p)/(q - p) = (32 + 2)/(32 - 2) = 34/30 = 17/15`
i.e. (q + p) : (q - p) = 17 :15
Case II:
When r = -2, x = -3
ab = x2 r = -18
cd = x2 r5 = -288
∴ `(q + p)/(q - p) = (-288 - 18)/(-288 + 18) = (-306)/(-270) = 17/15`
i.e., (q + p) : (q - p) = 17 : 15
Thus, in both the cases, we obtain (q+p) : (q − p) = 17 : 15
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