JEE MAIN - Mathematics (2024 - 9th April Morning Shift - No. 12)
Let $$\alpha, \beta$$ be the roots of the equation $$x^2+2 \sqrt{2} x-1=0$$. The quadratic equation, whose roots are $$\alpha^4+\beta^4$$ and $$\frac{1}{10}(\alpha^6+\beta^6)$$, is:
$$x^2-180 x+9506=0$$
$$x^2-195 x+9506=0$$
$$x^2-190 x+9466=0$$
$$x^2-195 x+9466=0$$
Explanation
$$\begin{aligned} & x^2+2 \sqrt{2 x}-1=0 \\ & \alpha+\beta=-2 \sqrt{2} \text { and } \alpha \beta=-1 \\ & \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta \\ & =8+2=10 \\ & \alpha^4+\beta^4=\left(\alpha^2+\beta^2\right)^2-2(\alpha \beta)^2 \\ & =100-2=98 \\ & \alpha^6+\beta^6=\left(\alpha^2+\beta^2\right)^3-3 \alpha^2 \beta^2\left(\alpha^2+\beta^2\right) \\ & =1000-3(10) \\ & =970 \\ & \therefore \quad \frac{1}{10}\left(\alpha^6+\beta^6\right)=97 \end{aligned}$$
Equation whose roots are $$\alpha^4+\beta^4$$ and $$\frac{1}{10}\left(\alpha^6+\beta^6\right)$$ is
$$\begin{aligned} & x^2-(98+97) x+98 \times 97=0 \\ & x^2-195 x+9506=0 \end{aligned}$$
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