If \(2x+\displaystyle\frac{9}{x}=9\), what is the minimum possible value of \(x^2+\displaystyle\frac{1}{x^2}\)?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{97}{36}\)
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 \(x\): \(2x^2+9=9x\), so \(2x^2-9x+9=0\)
\(2x^2-6x-3x+9=0\Rightarrow(2x-3)(x-3)=0\), so \(x=3\) or \(x=\displaystyle\frac{3}{2}\)
\(x=3\): \(x^2+\displaystyle\frac{1}{x^2}=9+\frac{1}{9}=\frac{82}{9}\)
\(x=\displaystyle\frac{3}{2}\): \(x^2+\displaystyle\frac{1}{x^2}=\frac{9}{4}+\frac{4}{9}=\frac{81+16}{36}=\frac{97}{36}\)
\(\displaystyle\frac{97}{36}\approx2.69\) is smaller than \(\displaystyle\frac{82}{9}\approx9.11\), so the minimum is \(\displaystyle\frac{97}{36}\)