Use the elimination method to determine whether the linear-quadratic system y = x^2 - 4x +10, y = -2x + 5 has zero, one, or two solutions.

Answer 1

To solve the linear-quadratic system using the elimination method, we'll start by equating the two expressions for y:

x^2 - 4x + 10 = -2x + 5

Next, we'll rearrange the equation to obtain a quadratic equation:

x^2 - 4x + 2x + 10 - 5 = 0

x^2 - 2x + 5 = 0

Now, we can solve this quadratic equation. However, we'll determine the number of solutions by calculating the discriminant first, which is given by:

b^2 - 4ac

In this case, a = 1, b = -2, and c = 5. So, the discriminant is:

(-2)^2 - 4(1)(5) = 4 - 20 = -16

The discriminant is negative, which means that the quadratic equation has no real solutions. Consequently, the linear-quadratic system y = x^2 - 4x + 10, y = -2x + 5 also has no real solutions.

Therefore, this system has zero solutions.