Question

The sum of the fourth powers of the roots of the equation \( 16x^2 - 10x + 1 = 0 \) is:

• A. \( \frac{257}{4096} \)
• B. \( \frac{257}{2048} \)
• C. \( \frac{257}{1024} \)
• D. \( \frac{257}{512} \)

Hint:

To compute higher powers of roots, use algebraic identities like: \[ \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha \beta)^2 \] and apply Vieta’s formulas carefully.

Answer 1

The Correct Option is A

Step 1: Let the roots of the quadratic be \( \alpha \) and \( \beta \) We are given: \[ 16x^2 - 10x + 1 = 0 \Rightarrow \text{Let roots be } \alpha, \beta \]
Step 2: Use identities for power sums We want: \[ \alpha^4 + \beta^4 \] Use identity: \[ \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2\alpha^2\beta^2 \] We first need \( \alpha + \beta \), \( \alpha\beta \): From Vieta’s formulas: \[ \alpha + \beta = \frac{10}{16} = \frac{5}{8}, \quad \alpha \beta = \frac{1}{16} \] Now: \[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left( \frac{5}{8} \right)^2 - 2 \cdot \frac{1}{16} = \frac{25}{64} - \frac{1}{8} = \frac{25 - 8}{64} = \frac{17}{64} \] Also: \[ \alpha^2\beta^2 = (\alpha\beta)^2 = \left( \frac{1}{16} \right)^2 = \frac{1}{256} \] So: \[ \alpha^4 + \beta^4 = \left( \frac{17}{64} \right)^2 - 2 \cdot \frac{1}{256} = \frac{289}{4096} - \frac{2}{256} = \frac{289}{4096} - \frac{32}{4096} = \frac{257}{4096} \]