If x is rational and `4(x^(2)+(1)/(x^(2)))+16(x+(1)/(x))-57=0` , then the product of all possible values of x is

Options

A. 4
B. 3
C. 2
D. 1

Answer

To solve the equation \( 4\left(x^2 + \frac{1}{x^2}\right) + 16\left(x + \frac{1}{x}\right) - 57 = 0 \) and find the product of all possible rational values of \( x \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ 4\left(x^2 + \frac{1}{x^2}\right) + 16\left(x + \frac{1}{x}\right) - 57 = 0 \]

### Step 2: Use substitutions
Let \( y = x + \frac{1}{x} \). Then we can express \( x^2 + \frac{1}{x^2} \) in terms of \( y \):
\[ x^2 + \frac{1}{x^2} = y^2 - 2 \]
Now substitute this back into the equation:
\[ 4(y^2 - 2) + 16y - 57 = 0 \]

### Step 3: Simplify the equation
Distributing the 4 gives:
\[ 4y^2 - 8 + 16y - 57 = 0 \]
Combine like terms:
\[ 4y^2 + 16y - 65 = 0 \]

### Step 4: Solve the quadratic equation
Now we will use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 4, b = 16, c = -65 \):
\[ b^2 - 4ac = 16^2 - 4 \times 4 \times (-65) = 256 + 1040 = 1296 \]
Now calculate \( y \):
\[ y = \frac{-16 \pm \sqrt{1296}}{2 \times 4} = \frac{-16 \pm 36}{8} \]
Calculating the two possible values for \( y \):
1. \( y = \frac{20}{8} = 2.5 \)
2. \( y = \frac{-52}{8} = -6.5 \)

### Step 5: Find \( x \) from \( y \)
Now we need to find \( x \) from \( y \):
1. For \( y = 2.5 \):
\[ x + \frac{1}{x} = 2.5 \implies x^2 - 2.5x + 1 = 0 \]
Using the quadratic formula:
\[ x = \frac{2.5 \pm \sqrt{(2.5)^2 - 4 \cdot 1}}{2} = \frac{2.5 \pm \sqrt{6.25 - 4}}{2} = \frac{2.5 \pm \sqrt{2.25}}{2} = \frac{2.5 \pm 1.5}{2} \]
This gives:
- \( x = \frac{4}{2} = 2 \)
- \( x = \frac{1}{2} \)

2. For \( y = -6.5 \):
\[ x + \frac{1}{x} = -6.5 \implies x^2 + 6.5x + 1 = 0 \]
Using the quadratic formula:
\[ x = \frac{-6.5 \pm \sqrt{(6.5)^2 - 4 \cdot 1}}{2} = \frac{-6.5 \pm \sqrt{42.25 - 4}}{2} = \frac{-6.5 \pm \sqrt{38.25}}{2} \]
This gives two more values for \( x \).

### Step 6: Calculate the product of all possible values of \( x \)
The product of the roots of a quadratic \( ax^2 + bx + c = 0 \) is given by \( \frac{c}{a} \).

For \( y = 2.5 \):
- The product of roots \( = 1 \)

For \( y = -6.5 \):
- The product of roots \( = 1 \)

Thus, the total product of all possible values of \( x \) is:
\[ 2 \times \frac{1}{2} \times \text{(product from } y = -6.5\text{)} \]
Since both quadratics yield a product of 1, the overall product is:
\[ 1 \times 1 = 1 \]

### Final Answer
The product of all possible values of \( x \) is \( \boxed{1} \).