Step 1: Roots of the quadratic equation Given:
\[ x^2 + 2\sqrt{2}x - 1 = 0. \]
The sum of the roots:
\[ \alpha + \beta = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2} = -2\sqrt{2}. \]
The product of the roots:
\[ \alpha \beta = \frac{\text{constant term}}{\text{coefficient of } x^2} = -1. \]
Step 2: Compute \( \alpha^4 + \beta^4 \) Using the identity:
\[ \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha \beta)^2. \]
First, calculate \( \alpha^2 + \beta^2 \):
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta. \]
Substitute \( \alpha + \beta = -2\sqrt{2} \) and \( \alpha \beta = -1 \):
\[ \alpha^2 + \beta^2 = (-2\sqrt{2})^2 - 2(-1). \]
\[ \alpha^2 + \beta^2 = 8 + 2 = 10. \]
Now substitute into \( \alpha^4 + \beta^4 \):
\[ \alpha^4 + \beta^4 = (10)^2 - 2(-1)^2. \]
\[ \alpha^4 + \beta^4 = 100 - 2 = 98. \]
Step 3: Compute \( \alpha^6 + \beta^6 \) Using the identity:
\[ \alpha^6 + \beta^6 = (\alpha^3 + \beta^3)^2 - 2(\alpha^3 \beta^3). \]
First, calculate \( \alpha^3 + \beta^3 \) using:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)((\alpha^2 + \beta^2) - \alpha \beta). \]
Substitute \( \alpha + \beta = -2\sqrt{2} \), \( \alpha^2 + \beta^2 = 10 \), and \( \alpha \beta = -1 \):
\[ \alpha^3 + \beta^3 = (-2\sqrt{2})(10 - (-1)). \]
\[ \alpha^3 + \beta^3 = (-2\sqrt{2})(11) = -22\sqrt{2}. \]
Now calculate \( \alpha^3 \beta^3 \):
\[ \alpha^3 \beta^3 = (\alpha \beta)^3 = (-1)^3 = -1. \]
Substitute into \( \alpha^6 + \beta^6 \):
\[ \alpha^6 + \beta^6 = (-22\sqrt{2})^2 - 2(-1). \]
\[ \alpha^6 + \beta^6 = (484 \cdot 2) + 2 = 968 + 2 = 970. \]
Thus:
\[ \frac{1}{10}(\alpha^6 + \beta^6) = \frac{970}{10} = 97. \]
Step 4: Quadratic equation The roots of the quadratic equation are \( \alpha^4 + \beta^4 = 98 \) and:
\[ \frac{1}{10}(\alpha^6 + \beta^6) = 97. \]
The quadratic equation is:
\[ x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0. \]
Sum of roots:
\[ 98 + 97 = 195. \]
Product of roots:
\[ 98 \cdot 97 = 9506. \]
Thus, the quadratic equation is:
\[ x^2 - 195x + 9506 = 0. \]
Final Answer: Option (3).
If the roots of the quadratic equation \( ax^2 + bx + c = 0 \) are real and equal, then:
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: