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.

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.