MathematicsMedium129×since 2002Q5141Let λ≠0\lambda \ne 0λ=0 be in R. If α\alphaα and β\betaβ are the roots of the equation, x² - x + 2λ\lambdaλ = 0 and α\alphaα and γ\gammaγ are the roots of the equation, 3x2−10x+27λ=03{x^2} - 10x + 27\lambda = 03x2−10x+27λ=0, then βγλ{{\beta \gamma } \over \lambda }λβγ is equal to:A36B9C27D18Check answerSkip