If `a ,b ,c in R^+a n d2b=a+c ,` then check the nature of roots of equation `a x^2+2b x+c=0.`

Explanation

To determine the nature of the roots of the quadratic equation \( ax^2 + 2bx + c = 0 \) given that \( a, b, c \in \mathbb{R}^+ \) and \( 2b = a + c \), we can follow these steps:

### Step 1: Rewrite the condition
We start with the condition given in the problem:
\[ 2b = a + c \]
We can rearrange this to:
\[ a - 2b + c = 0 \]

### Step 2: Substitute \( x = -1 \)
We will substitute \( x = -1 \) into the quadratic equation:
\[ a(-1)^2 + 2b(-1) + c = 0 \]
This simplifies to:
\[ a - 2b + c = 0 \]
From Step 1, we see that this condition holds true.

### Step 3: Identify one root
Since substituting \( x = -1 \) satisfies the equation, we conclude that \( x = -1 \) is a root of the quadratic equation.

### Step 4: Use the product of roots
Let the other root be \( \beta \). By Vieta's formulas, the sum of the roots \( (-1 + \beta) \) is given by:
\[ - \frac{2b}{a} \]
Thus, we have:
\[ -1 + \beta = -\frac{2b}{a} \]
This leads to:
\[ \beta = -\frac{2b}{a} + 1 \]

### Step 5: Calculate the product of roots
The product of the roots is given by:
\[ (-1) \cdot \beta = \frac{c}{a} \]
Substituting for \( \beta \):
\[ - \beta = -\left(-\frac{2b}{a} + 1\right) = \frac{2b}{a} - 1 \]
Thus, we have:
\[ \frac{2b}{a} - 1 = \frac{c}{a} \]
Rearranging gives:
\[ \frac{2b - c}{a} = 1 \]
This means:
\[ 2b - c = a \]

### Step 6: Analyze the nature of roots
Since \( a, b, c \in \mathbb{R}^+ \), we know:
- \( a > 0 \)
- \( b > 0 \)
- \( c > 0 \)

From \( 2b = a + c \), we can see that \( 2b \) is the sum of two positive numbers \( a \) and \( c \), indicating that \( b \) is also positive.

### Conclusion
Since we have established that one root is \( -1 \) and the other root \( \beta \) is derived from positive values, we conclude that both roots are real.