If α, β, γ are the roots of the equation 2x3 – 3x2 + 6x + 1 = 0

1.	If α, β, γ are the roots of the equation 2x3 – 3x2 + 6x + 1 = 0, then α2 +β2+ γ2 is equal to(a) \(\frac{-15}{4}\)(b) \(\frac{-9}{4}\)(c) \(\frac{13}{4}\)(d) 4
Answer» (a) \(\frac{-15}{4}\) Given,α, β, γ are the roots of the 2x³ – 3x² + 6x + 1 = 0 Since S₁ = \(\frac{\text{Coefficient of}\,x^2}{\text{Coefficient of}\,x^3}\) = S₁= α + β + γ = \(-\big(\frac{-3}{2}\big)\) = \(\frac{3}{2}\) Since S₂ = \(\frac{\text{Coefficient of}\,x}{\text{Coefficient of}\,x^3}\) = S₂= αβ + βγ + αγ = \(\frac{6}{2}\) = 3 Since S₃ = \(\frac{\text{Coefficient of constant term}}{\text{Coefficient of}\,x^3}\) = S₃= αβγ = \(-\frac12\) Now, α² +β²+ γ²= (α +β + γ)² - 2(αβ + βγ + αγ) = \(\big(\frac{3}{2}\big)^2\) - 2 x 3 = \(\frac{9}{4}-6\) = \(\frac{-15}{4}\)

If α, β, γ are the roots of the equation 2x3 – 3x2 + 6x + 1 = 0, then α2 +β2+ γ2 is equal to(a) \(\frac{-15}{4}\)(b) \(\frac{-9}{4}\)(c) \(\frac{13}{4}\)(d) 4

Answer»

(a) \(\frac{-15}{4}\)

Given,α, β, γ are the roots of the 2x³ – 3x² + 6x + 1 = 0

Since S₁ = \(\frac{\text{Coefficient of}\,x^2}{\text{Coefficient of}\,x^3}\) = S₁= α + β + γ = \(-\big(\frac{-3}{2}\big)\) = \(\frac{3}{2}\)

Since S₂ = \(\frac{\text{Coefficient of}\,x}{\text{Coefficient of}\,x^3}\) = S₂= αβ + βγ + αγ = \(\frac{6}{2}\) = 3

Since S₃ = \(\frac{\text{Coefficient of constant term}}{\text{Coefficient of}\,x^3}\) = S₃= αβγ = \(-\frac12\)

Now, α² +β²+ γ²= (α +β + γ)² - 2(αβ + βγ + αγ)

= \(\big(\frac{3}{2}\big)^2\) - 2 x 3 = \(\frac{9}{4}-6\) = \(\frac{-15}{4}\)