The number of integral solutions of `2(x^(2)+(1)/(x^(2)))-7(x+(1)/(x)) +9=0` when `x != 0` is

Options

A. 1
B. 2
C. 4
D. 0

Answer

To solve the equation \( 2\left(x^2 + \frac{1}{x^2}\right) - 7\left(x + \frac{1}{x}\right) + 9 = 0 \) for the number of integral solutions when \( x \neq 0 \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 2\left(x^2 + \frac{1}{x^2}\right) - 7\left(x + \frac{1}{x}\right) + 9 = 0 \]
We can rewrite \( x^2 + \frac{1}{x^2} \) in terms of \( x + \frac{1}{x} \). We know that:
\[ x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2 \]
Let \( t = x + \frac{1}{x} \). Then:
\[ x^2 + \frac{1}{x^2} = t^2 - 2 \]
Substituting this into the original equation gives:
\[ 2(t^2 - 2) - 7t + 9 = 0 \]

### Step 2: Simplify the equation
Now we simplify the equation:
\[ 2t^2 - 4 - 7t + 9 = 0 \]
\[ 2t^2 - 7t + 5 = 0 \]

### Step 3: Factor the quadratic equation
Next, we factor the quadratic equation:
\[ 2t^2 - 7t + 5 = 0 \]
To factor, we can look for two numbers that multiply to \( 2 \times 5 = 10 \) and add up to \( -7 \). The numbers are \( -5 \) and \( -2 \). Thus, we rewrite the equation as:
\[ 2t^2 - 5t - 2t + 5 = 0 \]
Grouping gives:
\[ t(2t - 5) - 1(2t - 5) = 0 \]
Factoring out \( (2t - 5) \):
\[ (2t - 5)(t - 1) = 0 \]

### Step 4: Solve for \( t \)
Setting each factor to zero gives:
\[ 2t - 5 = 0 \quad \Rightarrow \quad t = \frac{5}{2} \]
\[ t - 1 = 0 \quad \Rightarrow \quad t = 1 \]

### Step 5: Solve for \( x \)
Now we substitute back to find \( x \):
1. For \( t = \frac{5}{2} \):
\[ x + \frac{1}{x} = \frac{5}{2} \]
Multiplying through by \( 2x \):
\[ 2x^2 - 5x + 2 = 0 \]
The discriminant \( D = (-5)^2 - 4 \cdot 2 \cdot 2 = 25 - 16 = 9 \) (which is positive), so there are 2 real solutions.

2. For \( t = 1 \):
\[ x + \frac{1}{x} = 1 \]
Multiplying through by \( x \):
\[ x^2 - x + 1 = 0 \]
The discriminant \( D = (-1)^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \) (which is negative), so there are no real solutions.

### Conclusion
The only valid case for integral solutions comes from the first equation, which yields 2 real solutions. Therefore, the number of integral solutions of the given equation is:

\[ \text{Number of integral solutions} = 2 \]