If `ax^(2)+bx+c=0`, `a ne 0`, `a`, `b`, `c in R` has distinct real roots in `(1,2)`, then `a` and `5a+2b+c` have (a) same sign (b) opposite sign (c) not determined (d) none of these

Options

A. same sign
B. opposite sign
C. not determined
D. none of these

Explanation

To solve the problem, we need to analyze the quadratic equation \( ax^2 + bx + c = 0 \) where \( a \neq 0 \) and \( a, b, c \in \mathbb{R} \). The roots of this equation are denoted as \( \alpha \) and \( \beta \), which are given to be distinct real roots lying in the interval (1, 2).

### Step 1: Identify the properties of the roots
Since \( \alpha \) and \( \beta \) are the roots of the quadratic equation, we can use Vieta's formulas:
- The sum of the roots:
\[ \alpha + \beta = -\frac{b}{a} \]
- The product of the roots:
\[ \alpha \beta = \frac{c}{a} \]

Given that both roots lie in the interval (1, 2), we have:
\[ 1 < \alpha < 2 \quad \text{and} \quad 1 < \beta < 2 \]

### Step 2: Analyze the sum and product of the roots
From the inequalities for the roots:
- Since \( \alpha > 1 \) and \( \beta > 1 \):
\[ \alpha + \beta > 2 \]
- Since \( \alpha < 2 \) and \( \beta < 2 \):
\[ \alpha + \beta < 4 \]
Thus, we can conclude:
\[ 2 < \alpha + \beta < 4 \]

### Step 3: Determine the sign of \( a \)
From the sum of the roots:
\[ -\frac{b}{a} = \alpha + \beta \]
Since \( 2 < \alpha + \beta < 4 \), we have:
\[ -\frac{b}{a} > 2 \quad \text{and} \quad -\frac{b}{a} < 4 \]
This implies:
\[ b < -2a \quad \text{and} \quad b > -4a \]
Thus, \( b \) can be either positive or negative depending on the sign of \( a \).

### Step 4: Analyze \( 5a + 2b + c \)
We need to evaluate \( 5a + 2b + c \). Using Vieta's formulas:
\[ c = a \alpha \beta \]
Substituting \( c \) into the expression gives:
\[ 5a + 2b + c = 5a + 2b + a \alpha \beta \]
We can factor out \( a \):
\[ = a(5 + 2\frac{b}{a} + \alpha \beta) \]
Now, substituting \( \frac{b}{a} = -(\alpha + \beta) \):
\[ = a(5 - 2(\alpha + \beta) + \alpha \beta) \]

### Step 5: Determine the sign of \( 5a + 2b + c \)
We need to analyze the expression \( 5 - 2(\alpha + \beta) + \alpha \beta \):
- Since \( 2 < \alpha + \beta < 4 \), we can substitute:
- For \( \alpha + \beta = 2 \):
\[ 5 - 2(2) + \alpha \beta = 5 - 4 + \alpha \beta = 1 + \alpha \beta \]
- For \( \alpha + \beta = 4 \):
\[ 5 - 2(4) + \alpha \beta = 5 - 8 + \alpha \beta = -3 + \alpha \beta \]

Given \( \alpha \beta \) is positive (since both roots are in (1, 2)), we can conclude that:
- The expression \( 5 - 2(\alpha + \beta) + \alpha \beta \) will vary depending on the values of \( \alpha \) and \( \beta \).

### Conclusion
Since we have established that:
- \( a \) and \( 5a + 2b + c \) can either be both positive or both negative depending on the values of \( \alpha \) and \( \beta \), we conclude that they have the same sign.

Thus, the correct answer is:
**(a) same sign.**