Question : If $2x+\frac{9}{x}=9$, what is the minimum value of $x^2+\frac{1}{x^2}$?
Option 1: $\frac{95}{36}$
Option 2: $\frac{97}{36}$
Option 3: $\frac{86}{25}$
Option 4: $\frac{623}{27}$
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Correct Answer: $\frac{97}{36}$
Solution : Given: $2x+\frac{9}{x}=9$ Multiplying both sides by $x$, we get, $⇒2x^2-9x+9=0$ $⇒(2x-3)(x-3)=0$ So, $x= \frac{3}{2},3$ Putting the values of $x$ in the equation, we get When $x=\frac{3}{2}$, $x^2+\frac{1}{x^2}=(\frac{3}{2})^2+\frac{1}{(\frac{3}{2})^2}=\frac{97}{36}=2.69$ When $x=3$, $x^2+\frac{1}{x^2}=3^2+\frac{1}{3^2}=\frac{82}{9}=9.11$ So, the minimum value is $\frac{97}{36}$. Hence, the correct answer is $\frac{97}{36}$.
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