If $\alpha$ and $\beta$ are two distinct negative roots of $x^5-5 x^3+5 x^2-1=0$, then the equation of least…

If $\alpha$ and $\beta$ are two distinct negative roots of $x^5-5 x^3+5 x^2-1=0$, then the equation of least degree with integer coefficients having $\sqrt{-\alpha}$ and $\sqrt{-\beta}$ as its roots is
  1. $x^2-3 x+1=0$
  2. $-x^4+5 x^2-5 x+1=0$
  3. $-x^4-5 x^2+5 x+1=0$
  4. $x^4-3 x^2+1=0$

Solution

$x^5-5 x^3+5 x^2-1=0$
By inspection $x=1$ is a root $\begin{aligned} & (x-1)\left(x^4+x^3-4 x^2+x+1\right)=0 \\ & \Rightarrow(x-1)^2\left(x^3+2 x^2-2 x-1\right) \doteq 0 \\ & \Rightarrow(x-1)^3\left(x^2+3 x+1\right)=0 \\ & \Rightarrow x=1, x=\frac{-3 \pm \sqrt{5}}{2} \\ & \therefore \alpha=\frac{-3+\sqrt{5}}{2}, \beta=\frac{-3-\sqrt{5}}{2} \end{aligned}$
Since imaginary roots occurs in pair. So, equation having roots $-i \sqrt{\alpha}, i \sqrt{\alpha}, i \sqrt{\beta},-i \sqrt{\beta}$ $\begin{aligned} & (x-i \sqrt{\alpha})(x+i \sqrt{\alpha})(x-i \sqrt{\beta})(x+i \sqrt{\beta})=0 \\ & \Rightarrow x^4+(\alpha+\beta) x^2+\alpha \beta=0 \Rightarrow x^4-3 x^2+1=0 \end{aligned}$

Asked in: AP EAMCET 2024 (22 May Shift 1)

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