Question : If $x^4+x^{-4}=47, x>0$, then what is the value of $x+\frac{1}{x}-2?$
Option 1: 1
Option 2: 0
Option 3: 5
Option 4: 3
Correct Answer: 1
Solution : Given expression, $x^4+x^{-4}=47$ ⇒ $x^4+\frac{1}{x^4}=47$ Adding 2 on both sides, ⇒ $x^4+\frac{1}{x^4}+2=47+2$ ⇒ $x^4+\frac{1}{x^4}+2\times x^2\times\frac{1}{x^2}=49$ ⇒ $(x^2+\frac{1}{x^2})^2=7^2$ [As $a^2+b^2+2ab=(a+b)^2$] ⇒ $x^2+\frac{1}{x^2}=7$ Again adding 2 on both sides, ⇒ $x^2+\frac{1}{x^2}+2=7+2$ ⇒ $x^2+\frac{1}{x^2}+2\times x\times\frac1x=9$ ⇒ $(x+\frac1x)^2=3^2$ ⇒ $x+\frac1x=3$ ⇒ $x+\frac1x-2=3-2$ $\therefore x+\frac1x-2=1$ Hence, the correct answer is 1.
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