What is the value of x at the intersection of `y = (8)/((x^(2) + 4))` = and x + y = 2 ?

Options

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

Answer

To find the value of \( x \) at the intersection of the equations \( y = \frac{8}{x^2 + 4} \) and \( x + y = 2 \), we will follow these steps:

### Step 1: Substitute \( y \) in the second equation
From the second equation, we can express \( y \) in terms of \( x \):
\[ y = 2 - x \]

### Step 2: Set the two equations for \( y \) equal to each other
Now, we can set the expression for \( y \) from the first equation equal to the expression for \( y \) from the second equation:
\[ 2 - x = \frac{8}{x^2 + 4} \]

### Step 3: Cross-multiply to eliminate the fraction
To eliminate the fraction, we cross-multiply:
\[ (2 - x)(x^2 + 4) = 8 \]

### Step 4: Expand the left side
Now, we will expand the left side:
\[ 2x^2 + 8 - x^3 - 4x = 8 \]

### Step 5: Simplify the equation
Next, we simplify the equation by moving all terms to one side:
\[ -x^3 + 2x^2 - 4x + 8 - 8 = 0 \]
This simplifies to:
\[ -x^3 + 2x^2 - 4x = 0 \]

### Step 6: Factor out common terms
We can factor out \( -x \):
\[ -x(x^2 - 2x + 4) = 0 \]

### Step 7: Solve for \( x \)
This gives us two cases to consider:
1. \( -x = 0 \) which implies \( x = 0 \)
2. Solve the quadratic \( x^2 - 2x + 4 = 0 \)

### Step 8: Use the quadratic formula on the second case
For the quadratic equation \( x^2 - 2x + 4 = 0 \), we use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -2, c = 4 \):
\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} \]
\[ x = \frac{2 \pm \sqrt{4 - 16}}{2} \]
\[ x = \frac{2 \pm \sqrt{-12}}{2} \]
\[ x = \frac{2 \pm 2i\sqrt{3}}{2} \]
\[ x = 1 \pm i\sqrt{3} \]

### Conclusion
Thus, the values of \( x \) at the intersection are \( x = 0 \) and \( x = 1 \pm i\sqrt{3} \). The real solution is \( x = 0 \), while the other two solutions are imaginary.

---