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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