If α, β, γ are the roots of x3 – 2x2 + 3x – 4 = 0, then the

1.	If α, β, γ are the roots of x3 – 2x2 + 3x – 4 = 0, then the value of α2 β2 + β2 γ2 + γ2 α2 is : (a) –7 (b) –5 (c) –3 (d) 0
Answer» (a) -7 Given, α, β, γ are the roots of x³ – 2x² + 3x – 4 = 0. Then, S₁ = α + β + γ = 2 S₂ = αβ + βγ + γα = 3 S₃ = αβγ = 4 Now, (αβ + βγ + γα)² = α²β² + β²γ² +γ²α² + 2αβ²γ + 2βγ²α + 2α²βγ = α²β² + β²γ² +γ²α² + 2αβγ (α + β + γ) ⇒ α²β² + β²γ² +γ²α² = (αβ + βγ + γα)² – 2αβγ (α + β + γ) = 3² – 2 × 4 × 2 = 9 – 16 = – 7.

If α, β, γ are the roots of x3 – 2x2 + 3x – 4 = 0, then the value of α2 β2 + β2 γ2 + γ2 α2 is : (a) –7 (b) –5 (c) –3 (d) 0

Answer»

(a) -7

Given, α, β, γ are the roots of x³ – 2x² + 3x – 4 = 0. Then,

S₁ = α + β + γ = 2

S₂ = αβ + βγ + γα = 3

S₃ = αβγ = 4

Now, (αβ + βγ + γα)² = α²β² + β²γ² +γ²α² + 2αβ²γ + 2βγ²α + 2α²βγ

= α²β² + β²γ² +γ²α² + 2αβγ (α + β + γ)

⇒ α²β² + β²γ² +γ²α² = (αβ + βγ + γα)² – 2αβγ (α + β + γ)

= 3² – 2 × 4 × 2 = 9 – 16 = – 7.