The possible values of n for which the equation `nx^(2)+(2n-1) x +(n-1)=0` has roots of opposite sign is/are by

Options

A. no value of n
B. all values of n
C. `-1 lt n lt 0`
D. `0 lt n lt 1`

Answer

To find the possible values of \( n \) for which the equation

\[ nx^2 + (2n - 1)x + (n - 1) = 0 \]

has roots of opposite signs, we can follow these steps:

### Step 1: Identify the coefficients
In the quadratic equation \( ax^2 + bx + c = 0 \), we have:
- \( a = n \)
- \( b = 2n - 1 \)
- \( c = n - 1 \)

### Step 2: Condition for roots to have opposite signs
For the roots to have opposite signs, the product of the roots \( \alpha \beta \) must be negative. According to Vieta's formulas, the product of the roots is given by:

\[ \alpha \beta = \frac{c}{a} = \frac{n - 1}{n} \]

We need this product to be less than zero:

\[ \frac{n - 1}{n} < 0 \]

### Step 3: Analyze the inequality
The fraction \( \frac{n - 1}{n} < 0 \) implies that the numerator and denominator must have opposite signs.

1. **Case 1**: \( n - 1 < 0 \) and \( n > 0 \)
- This gives \( n < 1 \) and \( n > 0 \), which implies \( 0 < n < 1 \).

2. **Case 2**: \( n - 1 > 0 \) and \( n < 0 \)
- This case is not possible since \( n - 1 > 0 \) implies \( n > 1 \), which contradicts \( n < 0 \).

Thus, the only valid case is \( 0 < n < 1 \).

### Step 4: Ensure the discriminant is positive
Next, we need to ensure that the discriminant of the quadratic equation is positive for real roots:

\[ D = b^2 - 4ac = (2n - 1)^2 - 4n(n - 1) \]

Calculating the discriminant:

\[ D = (2n - 1)^2 - 4n^2 + 4n \]
\[ = 4n^2 - 4n + 1 - 4n^2 + 4n \]
\[ = 1 \]

Since \( D = 1 > 0 \), the discriminant is always positive for any real \( n \).

### Conclusion
The possible values of \( n \) for which the equation has roots of opposite signs are:

\[ 0 < n < 1 \]