Question

If a and b are the roots of x² – 3x + p = 0 and c and d are the roots of x² – 12x + q = 0, where a, b, c, d from a GP, prove that (q + p): (q – p) = 17: 15.

Answer 1

Given data is, 

x² – 3x + p = 0 → (1) 

a and b are roots of (1) 

So, (x + a)(x + b) = 0 

x² - (a + b)x + ab = 0 

So, a + b = 3 and ab = p → (2) 

Given data is, x² – 12x + q = 0 → (3) 

c and d are roots of (1) So, (x + c)(x + d) = 0 

x² - (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, AR² , AR³ also forms a GP, with common ratio R. 

From (2), 

a + b = 3 

A + AR = 3

\(\frac{3}{A}\) = 1+ R → (5)

From (4), 

c + d = 12 

AR² + AR³ = 12 

AR² (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, AR² , and AR³ 

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.