If a and b are the roots of x2 – 3x + p = 0 and c, d are the roots x

1.	If a and b are the roots of x2 – 3x + p = 0 and c, d are the roots x2 – 12x + q = 0, where a, b, c, d form a G.P. Prove that (q + p) : (q – p) = 17 : 15.
Answer» Given that a and b are roots of x² – 3x + p = 0 ⇒ a + b = 3 and ab = p ...(i) It is given that c and d are roots of x² – 12x + q = 0 ⇒ c + d = 12 and cd = q...(ii) 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. ⇒ a = a, b = ar c = ar²d = ar³ Now, ∴a + b = 3 ⇒ a + ar = 3 ⇒ a(1 + r) = 3…(iii) c + d = 12 ⇒ ar² + ar³ = 12 ⇒ ar²(1 + r) = 12.....(iv) From (iii) and (iv) we get 3.r² = 12 ⇒ r² = 4 ⇒ r = ±2 Substituting the value of r in (iii) we get a = 1 ⇒ b = ar = 2 ∴ c = ar² = 2² = 4 d = ar³ = 2³ = 8 ⇒ ab = p = 2 and cd = 4×8 = 32 ⇒ q + p = 32 + 2 = 34 and q−p = 32−2 = 30 ⇒ q + p:q−p = 34:30 = 17:15 Hence, proved.

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

Answer»

Given that a and b are roots of x² – 3x + p = 0

⇒ a + b = 3 and ab = p ...(i)

It is given that c and d are roots of x² – 12x + q = 0

⇒ c + d = 12 and cd = q...(ii)

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.

⇒ a = a, b = ar c = ar²d = ar³

Now,

∴a + b = 3

⇒ a + ar = 3

⇒ a(1 + r) = 3…(iii)

c + d = 12

⇒ ar² + ar³ = 12

⇒ ar²(1 + r) = 12.....(iv)

From (iii) and (iv) we get

3.r² = 12

⇒ r² = 4

⇒ r = ±2

Substituting the value of r in (iii) we get a = 1

⇒ b = ar = 2

∴ c = ar² = 2² = 4

d = ar³ = 2³ = 8

⇒ ab = p = 2 and cd = 4×8 = 32

⇒ q + p = 32 + 2 = 34 and q−p = 32−2 = 30

⇒ q + p:q−p = 34:30 = 17:15

Hence, proved.