The sum of the fourth powers of the roots of the equation $16 \mathrm{x}^2-10 \mathrm{x}+1=0$ is
The sum of the fourth powers of the roots of the equation $16 \mathrm{x}^2-10 \mathrm{x}+1=0$ is
- $\frac{257}{4096}$
- $\frac{257}{2048}$
- $\frac{257}{1024}$
- $\frac{257}{512}$
Solution
$\begin{aligned} & \text {} 16 \mathrm{x}^2-10 \mathrm{x}+1=0 \\ & \Rightarrow 16 \mathrm{x}^2-8 \mathrm{x}-2 \mathrm{x}+1=0 \\ & \Rightarrow(2 \mathrm{x}-1)(8 \mathrm{x}-1)=0 \\ & \Rightarrow \mathrm{x}=\frac{1}{2}, \frac{1}{8} \\ & \left(\left(\frac{1}{2}\right)^4+\left(\frac{1}{8}\right)^4\right)=\frac{1}{16}+\frac{1}{4096}=\frac{257}{4096}\end{aligned}$
Asked in: AP EAMCET 2023 (18 May Shift 2)
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