Question : If $(x^2-2x+1)=0$, then the value of $x^4+\frac{1}{x^4}$ is:
Option 1: 0
Option 2: 1
Option 3: 2
Option 4: 3
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Correct Answer: 2
Solution : Given: $(x^2-2x+1)=0$ ⇒ $x^2+1=2x$ Divide the given equation by $x$ on both sides, we get, ⇒ $x+\frac{1}{x}=2$ On squaring the above equation on both sides, we get, ⇒ $x^2+\frac{1}{x^2}+2=4$ ⇒ $x^2+\frac{1}{x^2}=2$ On squaring the above equation on both sides, we get, ⇒ $(x^2+\frac{1}{x^2})^2=2^2$ ⇒ $(x^4+\frac{1}{x^4})+2=4$ ⇒ $(x^4+\frac{1}{x^4})=2$ Hence, the correct answer is 2.
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Question : If $2x+\frac{2}{x}=3$, then the value of $x^{3}+\frac{1}{x^{3}}+2$ is:
Question : If $(x+\frac{1}{x})\neq 0$ and $(x^3+\frac{1}{x^3})= 0$, then the value $(x+\frac{1}{x})^4$ is:
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