If the sum of the squares of the reciprocals of the roots α and β of the equation 3 x2+λ x-1=0 is 15 , then 6(α3+β3)2 is equal to :

Question

If the sum of the squares of the reciprocals of the roots α and β of the equation 3 x2+λ x-1=0 is 15 , then 6(α3+β3)2 is equal to : (A) 18 (B) 24 (C) 36 (D) 96

Tardigrade Admin · Accepted Answer

Here α, β roots of equation 3 x2+λ x-1=0 α+β=(-λ/3), α β=(-1/3) (1/α2)+(1/β2)=((α+β)2-2 α β/α2 β2)=15 λ2=9 Now 6(α3+β3)2=6((α+β)((α+β)2-3 α β))2 =6((λ2/9)) (λ2/9)+1 2=24

Q. If the sum of the squares of the reciprocals of the roots α and β of the equation 3x2+λx−1=0 is 15 , then 6(α3+β3)2 is equal to :

18

24

36

96

Here α,β roots of equation 3x2+λx−1=0 α+β=3−λ​,αβ=3−1​ α21​+β21​=α2β2(α+β)2−2αβ​=15 λ2=9 Now 6(α3+β3)2=6((α+β)((α+β)2−3αβ))2 =6(9λ2​){9λ2​+1}2=24

Q. If the sum of the squares of the reciprocals of the roots $α$ and $β$ of the equation $3 x^{2} + λ x - 1 = 0$ is $15$ , then $6 (α^{3} + β^{3})^{2}$ is equal to :