The cubic equation whose roots are the square | vedclass

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(B) Let the roots of the equation $x^3-2x^2+3x-4=0$ be $\alpha, \beta, \gamma$. Then $\alpha+\beta+\gamma=2$,$\alpha\beta+\beta\gamma+\gamma\alpha=3$,

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$x^3+2x^2-7x-16=0$

Vedclass · Answer

(B) Let the roots of the equation $x^3-2x^2+3x-4=0$ be $\alpha, \beta, \gamma$. Then $\alpha+\beta+\gamma=2$,$\alpha\beta+\beta\gamma+\gamma\alpha=3$,and $\alpha\beta\gamma=4$. We want the equation whose roots are $\alpha^2, \beta^2, \gamma^2$. Let $y=x^2$,so $x=\sqrt{y}$. Substituting into the original equation: $(\sqrt{y})^3-2(\sqrt{y})^2+3\sqrt{y}-4=0$. $y\sqrt{y}-2y+3\sqrt{y}-4=0$. $\sqrt{y}(y+3)=2y+4$. Squaring both sides: $y(y+3)^2=(2y+4)^2$. $y(y^2+6y+9)=4y^2+16y+16$. $y^3+6y^2+9y=4y^2+16y+16$. $y^3+2y^2-7y-16=0$. Thus,the required equation is $x^3+2x^2-7x-16=0$.

The cubic equation whose roots are the squares of the roots of the equation $x^3-2x^2+3x-4=0$ is

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