QJEEM24Mafbcf: JEE Main 2024 Mathematics - Class 11 JEEM24Mafbcf Solution: Let be the roots of the equation . The quadratic equation,…

$\begin{aligned}& x^2+2 \sqrt{2 x}-1=0 \\& \alpha+\beta=-2 \sqrt{2} \text { and } \alpha \beta=-1 \\& \alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta \\& =8+2=10 \\& \alpha^4+\beta^4=\left(\alpha^2+\beta^2\right)^2-2(\alpha \beta)^2 \\& =100-2=98 \\& \alpha^6+\beta^6=\left(\alpha^2+\beta^2\right)^3-3 \alpha^2 \beta^2\left(\alpha^2+\beta^2\right) \\& =1000-3(10) \\& =970 \\& \therefore \quad \frac{1}{10}\left(\alpha^6+\beta^6\right)=97\end{aligned}$

Equation whose roots are $\alpha^4+\beta^4$ and $\frac{1}{10}\left(\alpha^6+\beta^6\right)$ is

$\begin{aligned}& x^2-(98+97) x+98 \times 97=0 \\& x^2-195 x+9506=0\end{aligned}$