If `2x + ((9)/(x)) =9,` what is the minimum vlaue of `x ^(2) + ((1)/( x ^(2)))` ?

Options

A. `95/36`
B. `97/36`
C. `86/25`
D. `(623)/(27)`

Answer

To solve the equation \(2x + \frac{9}{x} = 9\) and find the minimum value of \(x^2 + \frac{1}{x^2}\), we will follow these steps:

### Step 1: Rearranging the Equation
Start with the equation:
\[ 2x + \frac{9}{x} = 9 \]
Subtract 9 from both sides:
\[ 2x + \frac{9}{x} - 9 = 0 \]
This simplifies to:
\[ 2x + \frac{9}{x} - 9 = 0 \]

### Step 2: Multiplying through by \(x\)
To eliminate the fraction, multiply the entire equation by \(x\) (assuming \(x \neq 0\)):
\[ 2x^2 + 9 - 9x = 0 \]
Rearranging gives:
\[ 2x^2 - 9x + 9 = 0 \]

### Step 3: Using the Quadratic Formula
Now we can use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 2\), \(b = -9\), and \(c = 9\):
\[ b^2 - 4ac = (-9)^2 - 4 \cdot 2 \cdot 9 = 81 - 72 = 9 \]
Now substituting back into the formula:
\[ x = \frac{9 \pm \sqrt{9}}{2 \cdot 2} = \frac{9 \pm 3}{4} \]
This gives us two possible values for \(x\):
\[ x = \frac{12}{4} = 3 \quad \text{and} \quad x = \frac{6}{4} = \frac{3}{2} \]

### Step 4: Finding \(x^2 + \frac{1}{x^2}\)
Now we need to find the minimum value of \(x^2 + \frac{1}{x^2}\). We will calculate this for both values of \(x\):
1. For \(x = 3\):
\[ x^2 + \frac{1}{x^2} = 3^2 + \frac{1}{3^2} = 9 + \frac{1}{9} = 9 + 0.1111 = 9.1111 \]
2. For \(x = \frac{3}{2}\):
\[ x^2 + \frac{1}{x^2} = \left(\frac{3}{2}\right)^2 + \frac{1}{\left(\frac{3}{2}\right)^2} = \frac{9}{4} + \frac{4}{9} \]
To add these fractions, we need a common denominator:
\[ \text{LCM of 4 and 9 is 36} \]
Thus:
\[ \frac{9}{4} = \frac{81}{36} \quad \text{and} \quad \frac{4}{9} = \frac{16}{36} \]
Adding these gives:
\[ x^2 + \frac{1}{x^2} = \frac{81}{36} + \frac{16}{36} = \frac{97}{36} \]

### Step 5: Conclusion
Now we compare the two values:
- For \(x = 3\), \(x^2 + \frac{1}{x^2} = 9.1111\)
- For \(x = \frac{3}{2}\), \(x^2 + \frac{1}{x^2} = \frac{97}{36} \approx 2.6944\)

Thus, the minimum value of \(x^2 + \frac{1}{x^2}\) is:
\[ \frac{97}{36} \]