Question

Use the substitution method to determine whether the linear-quadratic system y=x^2−2x−6 , y=−5x−9 has 0, 1, or 2 solutions.(1 point) This system has solution(s).

Answer 1

To use the substitution method to solve the given linear-quadratic system:

1. **Equations:**
- \( y = x^2 - 2x - 6 \) (quadratic)
- \( y = -5x - 9 \) (linear)

2. **Substitute the linear equation into the quadratic equation:**
We set the two equations equal to each other since they both equal \( y \):
\[
-5x - 9 = x^2 - 2x - 6
\]

3. **Rearrange the equation:**
Move all terms to one side:
\[
0 = x^2 - 2x + 5x - 6 + 9
\]
Combine like terms:
\[
0 = x^2 + 3x + 3
\]
This can be rewritten as:
\[
x^2 + 3x + 3 = 0
\]

4. **Find the discriminant:**
The discriminant \( D \) of a quadratic \( ax^2 + bx + c \) is given by \( D = b^2 - 4ac \). For our equation:
- \( a = 1 \)
- \( b = 3 \)
- \( c = 3 \)

Calculate the discriminant:
\[
D = 3^2 - 4 \cdot 1 \cdot 3 = 9 - 12 = -3
\]

5. **Determine the number of solutions:**
Since the discriminant \( D \) is negative, this means the quadratic equation has no real solutions. Thus, there are no points where the linear and quadratic functions intersect.

### Conclusion
The system has **0 solutions**.