Question

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 use the elimination method, let's set the two given equations equal to each other:

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

Rearranging this equation, we get:

x^2 - 2x + 5 = 0

Now we have a quadratic equation. We can solve this by factoring or using the quadratic formula. In this case, the quadratic equation does not factor nicely, so we will use the quadratic formula:

x = (-b ± √(b^2 - 4ac))/2a

In our case, a = 1, b = -2, and c = 5. Substituting these values into the quadratic formula, we get:

x = (-(-2) ± √((-2)^2 - 4(1)(5)))/2(1)

Simplifying further, we have:

x = (2 ± √(4 - 20))/2

x = (2 ± √(-16))/2

Since the discriminant (√(-16)) is negative, the quadratic equation has no real solutions. Therefore, the linear-quadratic system y = x^2 - 4x + 10, y = -2x + 5 has no solutions.