Q. 18

# If a and b are the roots of x^{2} – 3x + p = 0 and c, d are roots of x^{2} – 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 x^{2}−3x + p=0

∴ a + b = 3 and ab= p …(1)

[∵ __If α and β are roots of the equation ax ^{2} + bx + c=0 then α + β=−b/a and αβ=c/a__.]

It is given that c and d are roots of x^{2}−12x + q=0

∴ c + d = 12 and cd=q …(2)

[∵ __If α and β are roots of the equation ax ^{2} + 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 = ar^{2}

d = ar^{3}

Now,

L.H.S

Now, From (1)

a + b=3

⇒ a + ar=3

⇒ a(1 + r)=3 ...(3)

From (2),

c + d=12

⇒ ar^{2} + ar^{3}=12

⇒ ar^{2}(1 + r)=12 ...(4)

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

r^{2} = 4

∴ r^{4} = 16

putting the value of r^{4} in L.H.S, we get -

Hence proved.

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