The equation `(x^(2)+3x+4)^(2)+3(x^(2)+3x+4)+4=x` has

Options

A. all its solutions real but not all positive
B. only two of its solutions real
C. two of its solutions positive and negative
D. none of solutions real

Explanation

To solve the equation \((x^2 + 3x + 4)^2 + 3(x^2 + 3x + 4) + 4 = x\), we will follow these steps:

### Step 1: Let \(y = x^2 + 3x + 4\)
We start by substituting \(y\) for \(x^2 + 3x + 4\). This simplifies our equation to:
\[ y^2 + 3y + 4 = x \]

### Step 2: Rearrange the equation
Rearranging gives us:
\[ y^2 + 3y + 4 - x = 0 \]

### Step 3: Analyze the quadratic in \(y\)
The equation \(y^2 + 3y + (4 - x) = 0\) is a quadratic equation in \(y\). To determine the nature of the solutions for \(y\), we need to calculate the discriminant \(D\):
\[ D = b^2 - 4ac = 3^2 - 4 \cdot 1 \cdot (4 - x) = 9 - 4(4 - x) = 9 - 16 + 4x = 4x - 7 \]

### Step 4: Determine the conditions for real solutions
For \(y\) to have real solutions, the discriminant must be non-negative:
\[ 4x - 7 \geq 0 \]
This simplifies to:
\[ 4x \geq 7 \quad \Rightarrow \quad x \geq \frac{7}{4} \]

### Step 5: Analyze the quadratic in \(x\)
Now we need to check the nature of the solutions for \(x\). The original equation can be rewritten as:
\[ y^2 + 3y + 4 - x = 0 \]
Substituting back \(y = x^2 + 3x + 4\):
\[ (x^2 + 3x + 4)^2 + 3(x^2 + 3x + 4) + 4 - x = 0 \]

### Step 6: Check the discriminant for \(x\)
To analyze the nature of the solutions for \(x\), we need to consider the discriminant of the quadratic formed in \(x\). However, since we already have the condition \(x \geq \frac{7}{4}\), we can conclude that there will be solutions for \(x\) in this range.

### Final Conclusion
The equation \((x^2 + 3x + 4)^2 + 3(x^2 + 3x + 4) + 4 = x\) has real solutions for \(x \geq \frac{7}{4}\).

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