Let alpha, beta be the roots of the equation x^2+2√2x-1 the quadratic equation whose roots are a^4+B^4 and 1/10(a^6+B^6)is

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?

---

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.

Learn how to find the quadratic equation whose roots are α⁴ + β⁴ and 1/10(α⁶ + β⁶). This problem involves quadratic equations, symmetric sums of powers, and Vieta's formulas. Suitable for high school students studying advanced algebra.

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