If alpha and beta are two distinct negative r | vedclass

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(D) Given equation: $x^5-5x^3+5x^2-1=0$. By inspection,$x=1$ is a root. Factoring the polynomial: $(x-1)(x^4+x^3-4x^2+x+1)=0$. Further factoring: $(x-

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$x^4-3x^2+1=0$

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(D) Given equation: $x^5-5x^3+5x^2-1=0$. By inspection,$x=1$ is a root. Factoring the polynomial: $(x-1)(x^4+x^3-4x^2+x+1)=0$. Further factoring: $(x-1)^2(x^3+2x^2-2x-1)=0$. $(x-1)^3(x^2+3x+1)=0$. The roots are $x=1$ (triple root) and $x=\frac{-3 \pm \sqrt{5}}{2}$. The two distinct negative roots are $\alpha = \frac{-3+\sqrt{5}}{2}$ and $\beta = \frac{-3-\sqrt{5}}{2}$. We need an equation with roots $\sqrt{-\alpha}$ and $\sqrt{-\beta}$. Let $y = \sqrt{-\alpha} \Rightarrow y^2 = -\alpha$. Since $\alpha$ and $\beta$ are roots of $x^2+3x+1=0$,we have $\alpha+\beta = -3$ and $\alpha\beta = 1$. Let the new roots be $y_1 = \sqrt{-\alpha}$ and $y_2 = \sqrt{-\beta}$. The equation is $(y^2+\alpha)(y^2+\beta) = 0$. $y^4 + (\alpha+\beta)y^2 + \alpha\beta = 0$. Substituting the values: $y^4 - 3y^2 + 1 = 0$.

If $\alpha$ and $\beta$ are two distinct negative roots of $x^5-5x^3+5x^2-1=0$,then the equation of least degree with integer coefficients having $\sqrt{-\alpha}$ and $\sqrt{-\beta}$ as its roots is

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