1. Given quadratic and basic symmetric sums

The roots $\alpha,\beta$ satisfy $x^2+2\sqrt{2}x-1=0.$

So, $\alpha+\beta=-2\sqrt{2}, \qquad \alpha\beta=-1.$

We need the quadratic whose roots are $r_1=\alpha^4+\beta^4, \qquad r_2=\frac{1}{10}(\alpha^6+\beta^6).$

So we must find:

• $r_1+r_2$
• $r_1r_2$

2. Compute $\alpha^2+\beta^2$

Using $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta,$ we get $\alpha^2+\beta^2=(-2\sqrt{2})^2-2(-1)=8+2=10.$

3. Compute $\alpha^4+\beta^4$

Use $\alpha^4+\beta^4=(\alpha^2+\beta^2)^2-2\alpha^2\beta^2.$ Since $\alpha^2\beta^2=(\alpha\beta)^2=1,$ we get $\alpha^4+\beta^4=10^2-2(1)=100-2=98.$

Thus, $r_1=98.$

4. Compute $\alpha^3+\beta^3$

Use $\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta).$ So, $\alpha^3+\beta^3=(-2\sqrt{2})^3-3(-1)(-2\sqrt{2}).$ Now, $(-2\sqrt{2})^3=-16\sqrt{2},$ therefore $\alpha^3+\beta^3=-16\sqrt{2}-6\sqrt{2}=-22\sqrt{2}.$

5. Compute $\alpha^6+\beta^6$

Use $\alpha^6+\beta^6=(\alpha^3+\beta^3)^2-2\alpha^3\beta^3.$ Now, $\alpha^3\beta^3=(\alpha\beta)^3=(-1)^3=-1.$ Hence, $\alpha^6+\beta^6=(-22\sqrt{2})^2-2(-1).$ Since $(-22\sqrt{2})^2=484\cdot 2=968,$ we get $\alpha^6+\beta^6=968+2=970.$

Thus, $r_2=\frac{1}{10}(\alpha^6+\beta^6)=\frac{970}{10}=97.$

6. Form the required quadratic

The required roots are $98$ and $97$.

So the quadratic equation is $x^2-(98+97)x+(98)(97)=0.$

Now, $98+97=195,$ and $98\cdot 97=9506.$

Therefore the equation is $x^2-195x+9506=0.$

7. Check with options

This matches: Option B: $x^2-195x+9506=0$.

8. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.