Question

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) = 17:15.

Answer 1

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
$xr^2\frac{ (1 + r) }{ x(1 + r)} = \frac{12 }{ 3}$
⇒ r2 = 4
⇒ r = ±2
When $r = 2 , x =\frac{3 }{ 1 + 2} =\frac{ 3 }{ 3} = 1$
When $r = -2 , x = \frac{3 }{1 - 2} = \frac{3 }{- 1} = - 3$
Case I:
When r = 2 and x =1,
ab = x 2 r = 2
cd = x 2 r 5 = 32
$∴\,\frac{ q + p }{ q - p} = \frac{32 + 2 }{ 32 - 2} = \frac{34 }{ 30} = \frac{17 }{15}$
i.e., (q + p) : (q - p) = 17 : 15
Case II:
When r = -2, x = -3,
ab = x 2 r = -18
cd = x 2 r 5 = -288
$∴ \frac{q + p }{ q - p} = \frac{- 288 - 18 }{ -288 + 18} = \frac{- 306 }{ - 270 }= \frac{17 }{ 15}$
i.e., (q + p) : (q - p) = 17 : 15
Thus, in both the cases, we obtain (q + p): (q- p) = 17:15.