If \(x^4+\displaystyle\frac{1}{x^4}=20162\), what is the positive value of \(x+\displaystyle\frac{1}{x}\)?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 12
Work back one step at a time with \(\left(t+\frac{1}{t}\right)^2=t^2+\frac{1}{t^2}+2\): first with \(t=x^2\), then with \(t=x\).
\(\left(x^2+\displaystyle\frac{1}{x^2}\right)^2=x^4+\frac{1}{x^4}+2=20164\)
\(x^2+\displaystyle\frac{1}{x^2}\) is positive, so \(x^2+\displaystyle\frac{1}{x^2}=\sqrt{20164}=142\)
\(\left(x+\displaystyle\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}+2=144\)
Positive value: \(x+\displaystyle\frac{1}{x}=\sqrt{144}=12\)