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:

• A. \( x^2 - 190x + 9466 = 0 \)
• B. \( x^2 - 195x + 9466 = 0 \)
• C. \( x^2 - 195x + 9506 = 0 \)
• D. \( x^2 - 180x + 9506 = 0 \)

Answer 1

The Correct Option is C

To find the quadratic equation whose roots are \(\alpha^4 + \beta^4\) and \(\frac{1}{10}(\alpha^6 + \beta^6)\), we start by analyzing the roots \(\alpha\) and \(\beta\) given by the equation \(x^2 + 2\sqrt{2}x - 1 = 0\).

From the given quadratic equation, we know:

• The sum of roots \((\alpha + \beta) = -2\sqrt{2}\).
• The product of roots \((\alpha \beta) = -1\).

Using these, we can find powers of \(\alpha\) and \(\beta\):

Finding \(\alpha^4 + \beta^4\)

First, compute \(\alpha^2 + \beta^2\):

\((\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta = 8\)
\(\Rightarrow \alpha^2 + \beta^2 = 8 - 2(-1) = 8 + 2 = 10\)

Then, calculate \(\alpha^4 + \beta^4\) using:

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

Finding \(\alpha^6 + \beta^6\)

We can find \(\alpha^6 + \beta^6\) using the identity:

\(\alpha^6 + \beta^6 = (\alpha^2 + \beta^2)(\alpha^4 + \beta^4) - \alpha^2\beta^2(\alpha^2 + \beta^2)\)
\(= 10 \times 98 - 1 \times 10 = 980 - 10 = 970\)

Now, remember the requirement for the quadratic roots:

\(x = \alpha^4 + \beta^4 = 98\)
\(y = \frac{1}{10}(\alpha^6 + \beta^6) = \frac{970}{10} = 97\)

The quadratic equation formed by these roots is:

• Sum of roots: \(98 + 97 = 195\)
• Product of roots: \(98 \times 97 = 9506\)

The quadratic equation is thus:

\(x^2 - 195x + 9506 = 0\)

Therefore, the correct option is \( x^2 - 195x + 9506 = 0 \).

Answer 2

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).