Mathematics - Quadratic Equation Question with Solution | TestHub

$x^2-4\sqrt{2}kx+2e^{4\ln k}-1=0$ or, $x^2-4\sqrt{2}kx+2k^4-1=0$ $\alpha +\beta =4\sqrt{2}k$ and $\alpha \cdot \beta =2k^4-1$ Squaring both sides, we get $$\begin{array}{rl}& (\alpha +\beta {)}^2=(4\sqrt{2}k{)}^2\\ & \Rightarrow {\alpha }^2+{\beta }^2+2\alpha \beta =32k^2\\ & 66+2\alpha \beta =32k^2\\ & 66+2\left(2k^4-1\right)=32k^2\\ & 66+4k^4-2=32k^2\\ & \Rightarrow 4k^4-32k^2+64=0\\ & \text{ or, }{k}^4-8k^2+16=0\\ & \Rightarrow {\left(k^2\right)}^2-8k^2+16=0\\ & \Rightarrow \left(k^2-4\right)\left(k^2-4\right)=0\\ & \Rightarrow k^2=4,k^2=4\\ & \Rightarrow k=\pm 2\end{array}$$ Now, ${\alpha }^3+{\beta }^3=(\alpha +\beta )\left({\alpha }^2+{\beta }^2-\alpha \beta \right)$ $$\therefore {\alpha }^3+{\beta }^3=(4\sqrt{2}k)\left[66-\left(2k^4-1\right)\right]$$ Putting $k=-2,(k=+2$ cannot be taken because it does not satisfy the above equation) $\therefore {\alpha }^3+{\beta }^3=(4\sqrt{2}(-2))\left[66-2(-2{)}^4-1\right]$ ${\alpha }^3+{\beta }^3=(-8\sqrt{2})(66-32+1)=$ $(-8\sqrt{2})(35)$ $\therefore {\alpha }^3+{\beta }^3=-280\sqrt{2}$