The cubic equation whose roots are the square | vedclass

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Question

(C) Let $\alpha, \beta, \gamma$ be the roots of the equation $12x^3-20x^2+x+3=0$.From Vieta's formulas:$\alpha+\beta+\gamma = \frac{20}{12} = \frac{5}

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$144x^3-376x^2+121x-9=0$

Vedclass · Answer

(C) Let $\alpha, \beta, \gamma$ be the roots of the equation $12x^3-20x^2+x+3=0$.From Vieta's formulas:$\alpha+\beta+\gamma = \frac{20}{12} = \frac{5}{3}$$\alpha\beta+\beta\gamma+\gamma\alpha = \frac{1}{12}$$\alpha\beta\gamma = -\frac{3}{12} = -\frac{1}{4}$We want the equation with roots $\alpha^2, \beta^2, \gamma^2$.Sum of roots: $\alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = (\frac{5}{3})^2 - 2(\frac{1}{12}) = \frac{25}{9} - \frac{1}{6} = \frac{50-3}{18} = \frac{47}{18} = \frac{376}{144}$.Sum of roots taken two at a time: $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2 = (\alpha\beta+\beta\gamma+\gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha+\beta+\gamma) = (\frac{1}{12})^2 - 2(-\frac{1}{4})(\frac{5}{3}) = \frac{1}{144} + \frac{5}{6} = \frac{1+120}{144} = \frac{121}{144}$.Product of roots: $\alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = (-\frac{1}{4})^2 = \frac{1}{16} = \frac{9}{144}$.The required equation is $x^3 - (\sum \alpha^2)x^2 + (\sum \alpha^2\beta^2)x - (\alpha^2\beta^2\gamma^2) = 0$.Substituting the values: $x^3 - \frac{376}{144}x^2 + \frac{121}{144}x - \frac{9}{144} = 0$.Multiplying by $144$: $144x^3 - 376x^2 + 121x - 9 = 0$.

The cubic equation whose roots are the squares of the roots of the equation $12x^3-20x^2+x+3=0$ is

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