If a,b,c are positive real numbers, then the number of positive real roots of the equation `ax^(2)+bx+c=0` is

Options

A. are real and positive
B. real and negative
C. have negative real part
D. have positive real part.

Explanation

To determine the number of positive real roots of the quadratic equation \( ax^2 + bx + c = 0 \) where \( a, b, c \) are positive real numbers, we can follow these steps:

### Step 1: Identify the general form of the quadratic equation
The quadratic equation is given as \( ax^2 + bx + c = 0 \).

### Step 2: Calculate the discriminant
The discriminant \( D \) of the quadratic equation is given by:
\[ D = b^2 - 4ac \]
Since \( a, b, c > 0 \), we need to analyze the value of \( D \).

### Step 3: Analyze the discriminant
1. **If \( D < 0 \)**:
- The roots are complex (imaginary) and do not exist on the real number line.
- Since both roots are not real, there are **no positive real roots**.

2. **If \( D = 0 \)**:
- The equation has one real root (a repeated root).
- The root is given by:
\[ x = -\frac{b}{2a} \]
- Since \( a > 0 \) and \( b > 0 \), it follows that \( -\frac{b}{2a} < 0 \). Thus, this root is negative, and there are **no positive real roots**.

3. **If \( D > 0 \)**:
- The equation has two distinct real roots.
- The roots are given by:
\[ x = \frac{-b \pm \sqrt{D}}{2a} \]
- Again, since \( a > 0 \) and \( b > 0 \), both roots will be negative:
- The larger root (which is \( \frac{-b + \sqrt{D}}{2a} \)) will still be negative because \( \sqrt{D} < b \) (since \( D = b^2 - 4ac \) and \( 4ac > 0 \)).
- The smaller root (which is \( \frac{-b - \sqrt{D}}{2a} \)) will be even more negative.

### Conclusion
In all cases, whether the discriminant is negative, zero, or positive, the roots of the equation \( ax^2 + bx + c = 0 \) where \( a, b, c \) are positive real numbers will not be positive. Therefore, the number of positive real roots of the equation is:

**0 positive real roots.**