If \(3x+\displaystyle\frac{10}{x}=11\), what is the minimum possible value of \(x^2+\displaystyle\frac{1}{x^2}\)?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{706}{225}\)
The condition is a quadratic equation, so \(x\) has two possible values. The expression is worked out for each, and the smaller result is the minimum.
Multiplying by \(x\): \(3x^2+10=11x\), so \(3x^2-11x+10=0\)
\(3x^2-6x-5x+10=0\Rightarrow(3x-5)(x-2)=0\), so \(x=2\) or \(x=\displaystyle\frac{5}{3}\)
\(x=2\): \(x^2+\displaystyle\frac{1}{x^2}=4+\frac{1}{4}=\frac{17}{4}\)
\(x=\displaystyle\frac{5}{3}\): \(x^2+\displaystyle\frac{1}{x^2}=\frac{25}{9}+\frac{9}{25}=\frac{625+81}{225}=\frac{706}{225}\)
\(\displaystyle\frac{706}{225}\approx3.14\) is smaller than \(\displaystyle\frac{17}{4}=4.25\), so the minimum is \(\displaystyle\frac{706}{225}\)