Number of roots of `(x^(2)+5x+7)(-x^(2)+3x-4) = 0`

Options

A. (a) 0
B. (b) 2
C. (c) 4
D. (d) None of these

Answer

To solve the equation \((x^2 + 5x + 7)(-x^2 + 3x - 4) = 0\), we need to find the number of real roots. We can do this by analyzing each quadratic polynomial separately and checking their discriminants.

### Step 1: Analyze the first polynomial \(x^2 + 5x + 7\)

1. **Set the polynomial to zero**:
\[ x^2 + 5x + 7 = 0 \]

2. **Calculate the discriminant**:
The discriminant \(D\) for a quadratic equation \(ax^2 + bx + c = 0\) is given by:
\[ D = b^2 - 4ac \]
Here, \(a = 1\), \(b = 5\), and \(c = 7\):
\[ D = 5^2 - 4 \cdot 1 \cdot 7 = 25 - 28 = -3 \]

3. **Interpret the discriminant**:
Since \(D < 0\), the polynomial \(x^2 + 5x + 7\) has no real roots (only imaginary roots).

### Step 2: Analyze the second polynomial \(-x^2 + 3x - 4\)

1. **Set the polynomial to zero**:
\[ -x^2 + 3x - 4 = 0 \]
This can be rewritten as:
\[ x^2 - 3x + 4 = 0 \]

2. **Calculate the discriminant**:
Here, \(a = 1\), \(b = -3\), and \(c = 4\):
\[ D = (-3)^2 - 4 \cdot 1 \cdot 4 = 9 - 16 = -7 \]

3. **Interpret the discriminant**:
Since \(D < 0\), the polynomial \(-x^2 + 3x - 4\) also has no real roots (only imaginary roots).

### Step 3: Conclusion

Since both quadratic equations have no real roots, the product \((x^2 + 5x + 7)(-x^2 + 3x - 4) = 0\) has no real roots.

Thus, the number of real roots of the given equation is:
\[ \text{Number of real roots} = 0 \]

### Final Answer:
The correct answer is **Option A: 0**.

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