If \(x^4+\displaystyle\frac{1}{x^4}=1154\) and \(x>0\), then find the value of \(2(x-3)^2+30\).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 46
Work back with \(\left(t+\frac{1}{t}\right)^2=t^2+\frac{1}{t^2}+2\) to reach \(x+\frac{1}{x}\). That gives a quadratic in \(x\), and \((x-3)^2\) comes out of it by completing the square.
\(\left(x^2+\displaystyle\frac{1}{x^2}\right)^2=1154+2=1156\), so \(x^2+\displaystyle\frac{1}{x^2}=34\)
\(\left(x+\displaystyle\frac{1}{x}\right)^2=34+2=36\), and \(x>0\), so \(x+\displaystyle\frac{1}{x}=6\)
Multiplying by \(x\): \(x^2-6x+1=0\)
\(x^2-6x+9=8\), so \((x-3)^2=8\)
\(2(x-3)^2+30=2\times8+30=46\)