We need to find the quadratic equation whose roots are $\alpha^4 + \beta^4$ and $\frac{1}{10}(\alpha^6 + \beta^6)$, where $\alpha$ and $\beta$ are the roots of the equation:

$x^2 + 2\sqrt{2}x - 1 = 0$

Step 1: Find the roots $\alpha$ and $\beta$

The roots $\alpha$ and $\beta$ can be found using the quadratic formula:

$\alpha, \beta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 1$, $b = 2\sqrt{2}$, and $c = -1$. Plugging in these values:

$\alpha, \beta = \frac{-2\sqrt{2} \pm \sqrt{(2\sqrt{2})^2 - 4(1)(-1)}}{2(1)}$

$\alpha, \beta = \frac{-2\sqrt{2} \pm \sqrt{8 + 4}}{2}$

$\alpha, \beta = \frac{-2\sqrt{2} \pm \sqrt{12}}{2}$

$\alpha, \beta = \frac{-2\sqrt{2} \pm 2\sqrt{3}}{2}$

$\alpha, \beta = -\sqrt{2} \pm \sqrt{3}$

So the roots are:

$\alpha = -\sqrt{2} + \sqrt{3}, \quad \beta = -\sqrt{2} - \sqrt{3}$

Step 2: Calculate $\alpha^4 + \beta^4$

We will now compute $\alpha^4 + \beta^4$.

First, note that:

$\alpha + \beta = -2\sqrt{2}, \quad \alpha\beta = (\sqrt{2})^2 - (\sqrt{3})^2 = 2 - 3 = -1$

To find $\alpha^4 + \beta^4$, we use the identity:

$\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2\alpha^2\beta^2$

First, we need $\alpha^2 + \beta^2$ and $\alpha^2\beta^2$:

$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (-2\sqrt{2})^2 - 2(-1) = 8 + 2 = 10$

$\alpha^2\beta^2 = (\alpha\beta)^2 = (-1)^2 = 1$

Now:

$\alpha^4 + \beta^4 = (10)^2 - 2(1) = 100 - 2 = 98$

Step 3: Calculate $\frac{1}{10}(\alpha^6 + \beta^6)$

We use the identity:

$\alpha^6 + \beta^6 = (\alpha^2 + \beta^2)(\alpha^4 + \beta^4 - \alpha^2\beta^2)$

Using the values we computed:

$\alpha^6 + \beta^6 = (10)(98 - 1) = 10 \times 97 = 970$

Thus:

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

Step 4: Form the new quadratic equation

The roots of the desired quadratic equation are $98$ and $97$. Therefore, the equation is:

$x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$

The sum of the roots is $98 + 97 = 195$, and the product of the roots is $98 \times 97 = 9506$. So the quadratic equation is:

$x^2 - 195x + 9506 = 0$

This is the quadratic equation whose roots are $\alpha^4 + \beta^4$ and $\frac{1}{10}(\alpha^6 + \beta^6)$.

Would you like more details or have any questions?

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1. What is the general method for finding roots of a quadratic equation?
2. How can we derive identities like $\alpha^4 + \beta^4$ in terms of simpler expressions?
3. Why is the quadratic equation with roots $98$ and $97$ important here?
4. What are other identities related to symmetric sums of powers of roots?
5. How does the choice of signs in the quadratic formula affect the roots?

Tip: When solving quadratic equations, it's important to simplify intermediate expressions carefully to avoid mistakes in calculations.