If \(x^2+\displaystyle\frac{1}{x^2}=98\) and \(x>0\), then find the value of \(x^5+\displaystyle\frac{1}{x^5}\).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 95050
The fifth power comes from multiplying the second and third powers: \(\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=x^5+\frac{1}{x^5}+x+\frac{1}{x}\). So \(x+\frac{1}{x}\) and \(x^3+\frac{1}{x^3}\) are found first.
\(\left(x+\displaystyle\frac{1}{x}\right)^2=98+2=100\); since \(x>0\), \(x+\displaystyle\frac{1}{x}=10\)
\(x^3+\displaystyle\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)=1000-30=970\)
\(98\times970=x^5+\displaystyle\frac{1}{x^5}+10\)
\(x^5+\displaystyle\frac{1}{x^5}=95060-10=95050\)