# If a and b are the roots of x^2 – 3x + p = 0 and c, d are of x^2 – 12x + q = 0, where a, b, c, d form a G.P. Prove that (q + p) : (q – p) = 17:15.

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### 1 Answer

It is given that a and b are roots of x^2 – 3x + p = 0 and c, d are roots of x^2 – 12x + q = 0 such that a, b, c, d are in a GP. We have to show that (q + p) : (q – p) = 17:15.

Using Viete's formula for a quadratic polynomial of the form ax^2 + bx + c, the sum of the roots x1 + x2 = -b/a and the product x1*x2 = c/a

This gives a + b = 3 and a*b = p, c + d = 12 and c*d = q

As a, b, c, d are in a GP we can write b = ar, c = ar^2 and d = ar^3

a + ar = 3 and ar^2 + ar^3 = 12

(ar^2 + ar^3)/(a + ar) = 12/3

=> r^2 = 4

(q + p)/(q - p) = (c*d + a*b)/(c*d - a*b)

=> (ar^2*ar^3 + a*ar)/(ar^2*ar^3 - a*ar)

=> (r^4 + 1)/(r^4 - 1)

=> (16 + 1)/(16 - 1)

=> 17/15

**This proves (q + p) : (q – p) = 17:15.**