Let , be the roots of the equation x^2+2√2x-1=0.

Question

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:

Accepted Answer

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

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

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

Solutions · 1