If \(x^4+\displaystyle\frac{1}{x^4}=3842\), what is the positive value of \(x+\displaystyle\frac{1}{x}\)?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 8
Squaring \(t+\frac{1}{t}\) gives \(\left(t+\frac{1}{t}\right)^2=t^2+\frac{1}{t^2}+2\). Used twice, this steps down from the fourth powers to \(x+\frac{1}{x}\).
\(\left(x^2+\displaystyle\frac{1}{x^2}\right)^2=x^4+\frac{1}{x^4}+2=3842+2=3844\)
\(x^2+\displaystyle\frac{1}{x^2}=\sqrt{3844}=62\) (it is a sum of squares, so positive)
\(\left(x+\displaystyle\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}+2=62+2=64\)
Positive value: \(x+\displaystyle\frac{1}{x}=\sqrt{64}=8\)