Ans.(i)

Sol. Let roots are α 1 , α 2 , α 3

3 α 2 = b, α 1 α 2 + α 2 α 3 + α 3 α 1 = a

α 1 α 2 α 3 = 1, α 2 (2 α 2 ) + α 1 α 3 = a

α 1 α 3 = , 2 + α 1 α 3 = a

2 + – a = 0

2b

3 + 27 = 9ab

(ii)

Sol. a + b = – 1, ab = 2

⇒ f(1) = 1 – a + b – 1 = 0 ⇒ a = b

⇒ g(1) = 1 – b + a – 1 ⇒ a = b

⇒ f(1) + g(1) = 0 ⇒ 1 is root of f(x) + g(x) = 0

⇒ f(x) + g(x) = 2x 3 – (a + b) x 2 + (a + b) x –2

= 2x 3 + x 2 – x –2

= 2(x –1) (x 2 + x +1) + x(x –1)

= (x – 1) (2x 2 + 3x + 2)

⇒ one real root and two imaginary root

(iii)

Sol . x 3 = x 3 – ax 2 + bx – 1

ax 2 – bx + 1 = 0 ⇒ α + β = b/a, αβ = 1/a

α 2 + β 2 = αβ ⇒ – 2 = .

⇒ (–1 –a) 2 –2a = a

⇒ 3(a 2 + 2a +1) – 6a = 4a

⇒ 3a 2 – 4a + 3 = 0

⇒ a 1 + a 2 = , a 1 a 2 = 1

+ = ((a 1 + a 2 ) 2 – 3a 1 a 2 )

= = –

So, eqn. 27x 2 + 44x + 27 = 0