Let p, q, r `in` R and `r gt p gt 0`. If the quadratic equation `px^(2) + qx + r = 0` has two complex roots `alpha and beta`, then `|alpha|+|beta|`, is

Options

A. less than 2 but not equat to 1
B. equal to 2
C. equal to 1
D. greater than 2

Explanation

To solve the problem, we need to analyze the given quadratic equation and the conditions provided.

### Step-by-Step Solution:

1. **Identify the Quadratic Equation**:
The quadratic equation given is \( px^2 + qx + r = 0 \).

2. **Conditions on Roots**:
We know that the roots \( \alpha \) and \( \beta \) are complex. For a quadratic equation to have complex roots, the discriminant must be negative:
\[ D = q^2 - 4pr < 0 \]

3. **Roots in Conjugate Pairs**:
Since the roots are complex, they can be expressed as \( \alpha = a + bi \) and \( \beta = a - bi \), where \( a \) and \( b \) are real numbers, and \( b \neq 0 \). This means that \( |\alpha| = |\beta| \).

4. **Product of Roots**:
The product of the roots of the quadratic equation is given by:
\[ \alpha \beta = \frac{r}{p} \]
Since \( r > p > 0 \), we have:
\[ \frac{r}{p} > 1 \implies |\alpha| \cdot |\beta| > 1 \]

5. **Magnitude of Roots**:
Because \( |\alpha| = |\beta| \), we can let \( |\alpha| = |\beta| = k \). Therefore:
\[ k^2 = |\alpha| \cdot |\beta| > 1 \implies k > 1 \]

6. **Sum of Magnitudes**:
Now, we can find the sum of the magnitudes:
\[ |\alpha| + |\beta| = k + k = 2k \]
Since \( k > 1 \), it follows that:
\[ 2k > 2 \]

7. **Conclusion**:
Therefore, we conclude that:
\[ |\alpha| + |\beta| > 2 \]

### Final Answer:
The value of \( |\alpha| + |\beta| \) is greater than 2.

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