# If a and b are the roots of x² - 3x + p = 0 and c, d are the roots of x² -12x + q = 0 where a, b, c, d form a G.P. Prove that (q + p) : (q - p) = 17 : 15

**Solution:**

It is given that a and b are the roots of x^{2} - 3x + p = 0

Therefore,

a + b = 3, ab = p (by sum and product of roots of a quadratic equation)....(1)

Also, c and d are the roots of x^{2} - 12x + q = 0

c + d = 12, cd = q ....(2)

It is given that a,b, c, d form a G.P.

Let a = x, b = xr, c = xr^{2}, d = xr^{3}

Form (1) and (2) , we obtain

x + xr = 3 ⇒ x (1+ r) = 3 ....(3)

xr^{2} + xr^{3} = 12 ⇒ xr^{2} (1 + r) = 12 ....(4)

On dividing (4) by (3) , we obtain

[xr^{2} (1+ r)] / [x (1+ r)] = 12/3

⇒ r^{2} = 4

⇒ r = ± 2

__Case I__**: **

when r = 2, x = 1

ab = x^{2}r = 2 = p

cd = x^{2}r^{5} = 32 = q

Therefore,

⇒ (q + p)/(q - p) = (32 + 2)/(32 - 2) = 34/30 = 17/15

⇒ (q + p) : (q - p) = 17 : 15

__Case II__**: **when r = - 2, x = - 3

ab = x^{2}r = - 18 = p

cd = x^{2}r^{5} = - 288 = q

Therefore,

⇒ (q + p)/(q - p) = (- 288 - 18)/(- 288 + 18) = (- 306)/(- 270) = 17/15

⇒ (q + p) : (q - p) = 17 : 15

Thus, (q + p) : (q - p) = 17 : 15

Hence proved

NCERT Solutions Class 11 Maths Chapter 9 Exercise ME Question 18

## If a and b are the roots of x² - 3x + p = 0 and c, d are the roots of x² -12x + q = 0 where a, b, c, d form a G.P. Prove that (q + p) : (q - p) = 17 : 15

**Summary:**

If a and b are the roots of x² - 3x + p = 0 and c, d are the roots of x² -12x + q = 0 where a, b, c, d form a G.P, then we proved that (q + p) : (q - p) = 17 : 15

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