Roots of the equation `x^(2)+bx-c=0 (b,c gt 0)` are

Options

A. Both positive
B. Both negative
C. Of opposite sign
D. None of these

Answer

To find the roots of the quadratic equation \( x^2 + bx - c = 0 \), we can use the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In our case, the coefficients are:
- \( a = 1 \)
- \( b = b \) (the coefficient of \( x \))
- \( c = -c \) (the constant term)

Substituting these values into the quadratic formula gives us:

\[ x = \frac{-b \pm \sqrt{b^2 - 4(1)(-c)}}{2(1)} \]

This simplifies to:

\[ x = \frac{-b \pm \sqrt{b^2 + 4c}}{2} \]

Now, let's analyze the roots:

1. **Discriminant**: The discriminant \( D \) is given by \( D = b^2 + 4c \).
- Since \( b > 0 \) and \( c > 0 \), it follows that \( D > 0 \).
- This means the roots are real and distinct.

2. **Nature of Roots**:
- The roots can be expressed as:
\[ x_1 = \frac{-b + \sqrt{b^2 + 4c}}{2} \]
\[ x_2 = \frac{-b - \sqrt{b^2 + 4c}}{2} \]

3. **Sign of Roots**:
- For \( x_1 \):
\[ x_1 = \frac{-b + \sqrt{b^2 + 4c}}{2} \]
Since \( \sqrt{b^2 + 4c} > b \) (because \( 4c > 0 \)), we have \( -b + \sqrt{b^2 + 4c} > 0 \). Thus, \( x_1 > 0 \).

- For \( x_2 \):
\[ x_2 = \frac{-b - \sqrt{b^2 + 4c}}{2} \]
Here, both terms are negative, so \( x_2 < 0 \).

In conclusion, one root is positive and the other root is negative.

### Final Answer:
The roots of the equation \( x^2 + bx - c = 0 \) are one positive and one negative.