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 $\frac{1}{10} \left( \alpha^6 + \beta^6 \right)$, is:

Updated On: Apr 11, 2025
  • \( x^2 - 190x + 9466 = 0 \)
  • \( x^2 - 195x + 9466 = 0 \)
  • \( x^2 - 195x + 9506 = 0 \)
  • \( x^2 - 180x + 9506 = 0 \)
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The Correct Option is C

Solution and Explanation

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).

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