Question : If $x^2-\sqrt{7} x+1=0$, then what is the value of $x^5+\frac{1}{x^5} ?$
Option 1: $19 \sqrt{7}$
Option 2: $21 \sqrt{7}$
Option 3: $25 \sqrt{7}$
Option 4: $27 \sqrt{7}$
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Correct Answer: $19 \sqrt{7}$
Solution :
Given: $x^2-\sqrt{7} x+1=0$
Dividing by $x$, we get:
⇒ $x+\frac{1}{x}=\sqrt{7}$
Now, Squaring both sides, we get:
⇒ $(x+\frac{1}{x})^2=(\sqrt{7})^2$
⇒ $x^2+\frac{1}{x^2}+2=7$
⇒ $x^2+\frac{1}{x^2}=5$
Similarly, Cubing both sides, we get:
⇒ $(x+\frac{1}{x})^3=(\sqrt{7})^3$
⇒ $x^3+\frac{1}{x^3}+3(x\times\frac{1}{x})(x+\frac{1}{x})=7\sqrt7$
⇒ $x^3+\frac{1}{x^3}+3\sqrt 7=7\sqrt7$
⇒ $x^3+\frac{1}{x^3}=4\sqrt7$
Now, $x^5+\frac{1}{x^5}=(x^3+\frac{1}{x^3})\times(x^2+\frac{1}{x^2})-(x+\frac{1}{x})$
= $4\sqrt7\times5-\sqrt7$
= $20\sqrt7-\sqrt7$
= $19\sqrt7$
Hence, the correct answer is $19\sqrt7$.
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