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.