Q. 184.8( 4 Votes )

# If a and b are the roots of x^{2} – 3x + p = 0 and c and d are the roots of x^{2} – 12x + q = 0, where a, b, c, d from a GP, prove that (q + p): (q – p) = 17: 15.

Answer :

Given data is,

x^{2} – 3x + p = 0 → (1)

a and b are roots of (1)

So, (x + a)(x + b) = 0

x^{2} - (a + b)x + ab = 0

So, a + b = 3 and ab = p → (2)

Given data is,

x^{2} – 12x + q = 0 → (3)

c and d are roots of (1)

So, (x + c)(x + d) = 0

x^{2} - (c + d)x + cd = 0

So, c + d = 12 and cd = q → (4)

a, b, c, d are in GP.(Given data)

Similarly A, AR, AR^{2}, AR^{3} also forms a GP, with common ratio R.

From (2),

a + b = 3

A + AR = 3

→ (5)

From (4),

c + d = 12

AR^{2} + AR^{3} = 12

AR^{2} (1 + R) = 12 → (6)

Substituting value of (1 + R) in (6).

R = 2

Now, substitute value of R in (5) to get value of A,

A = 1

Now, the GP required is A, AR, AR^{2}, and AR^{3}

1, 2, 4, 8…is the required GP.

So,

a = 1, b = 2, c = 4, d = 8

From (2) and (4),

ab = p and cd = q

So, p = 2, and q = 32.

So, (q + p): (q – p) = 17: 15.

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