If the equation `(1+m)x^(2)-2(1+3m)x+(1-8m)=0` where `m epsilonR~{-1}`, has atleast one root is negative, then

Options

A. `m epsilon(-oo,-1)`
B. `m epsilon (-1/8,oo)`
C. `m epsilon(-1-1/8)`
D. `m epsilonR`

Answer

To solve the problem, we need to determine the values of \( m \) such that the quadratic equation

\[ (1+m)x^2 - 2(1+3m)x + (1-8m) = 0 \]

has at least one negative root.

### Step 1: Identify the coefficients
The given quadratic equation can be expressed in the standard form \( Ax^2 + Bx + C = 0 \), where:
- \( A = 1 + m \)
- \( B = -2(1 + 3m) \)
- \( C = 1 - 8m \)

### Step 2: Condition for at least one negative root
For the quadratic equation to have at least one negative root, we can use Vieta's formulas, which state:
- The sum of the roots \( \alpha + \beta = -\frac{B}{A} \)
- The product of the roots \( \alpha \beta = \frac{C}{A} \)

We need to ensure that at least one root is negative. This can happen in two scenarios:
1. One root is positive and the other is negative.
2. Both roots are negative.

### Step 3: Analyze the conditions
1. **Condition for the sum of roots:**
\[ \alpha + \beta = -\frac{B}{A} = \frac{2(1 + 3m)}{1 + m} \]
For at least one root to be negative, we need:
\[ \frac{2(1 + 3m)}{1 + m} < 0 \]
This inequality holds when the numerator and denominator have opposite signs.

2. **Condition for the product of roots:**
\[ \alpha \beta = \frac{C}{A} = \frac{1 - 8m}{1 + m} \]
For both roots to be negative, we need:
\[ \frac{1 - 8m}{1 + m} > 0 \]
This inequality holds when both the numerator and denominator are positive or both are negative.

### Step 4: Solve the inequalities
**For the sum of roots:**
1. \( 2(1 + 3m) > 0 \) and \( 1 + m < 0 \)
- \( 1 + 3m > 0 \) implies \( m > -\frac{1}{3} \)
- \( 1 + m < 0 \) implies \( m < -1 \) (not possible since \( m \) cannot be both greater than \(-\frac{1}{3}\) and less than \(-1\))

2. \( 2(1 + 3m) < 0 \) and \( 1 + m > 0 \)
- \( 1 + 3m < 0 \) implies \( m < -\frac{1}{3} \)
- \( 1 + m > 0 \) implies \( m > -1 \)

Thus, the valid range from this condition is:
\[ -1 < m < -\frac{1}{3} \]

**For the product of roots:**
1. \( 1 - 8m > 0 \) and \( 1 + m > 0 \)
- \( 1 - 8m > 0 \) implies \( m < \frac{1}{8} \)
- \( 1 + m > 0 \) implies \( m > -1 \)

Thus, the valid range from this condition is:
\[ -1 < m < \frac{1}{8} \]

### Step 5: Find the intersection of the ranges
From the conditions derived:
1. From the sum of roots: \( -1 < m < -\frac{1}{3} \)
2. From the product of roots: \( -1 < m < \frac{1}{8} \)

The intersection of these two ranges is:
\[ -1 < m < -\frac{1}{3} \]

### Final Answer
Thus, the values of \( m \) for which the quadratic equation has at least one negative root are:
\[ m \in (-1, -\frac{1}{3}) \]