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.

Enter 0 for zero solutions.

Enter 1 for one solution.

Enter 2 for two solutions.

(1 point)
This system has __ solution(s).

Answer 1

To determine whether the given linear-quadratic system has zero, one, or two solutions, we will substitute the linear equation into the quadratic equation and solve for \(x\).

The equations are:
1. \( y = x^2 - 4x + 10 \) (quadratic)
2. \( y = -2x + 5 \) (linear)

We can set the two equations equal since both equal \(y\):

\[
x^2 - 4x + 10 = -2x + 5
\]

Next, we rearrange the equation by moving all terms to one side:

\[
x^2 - 4x + 2x + 10 - 5 = 0
\]

This simplifies to:

\[
x^2 - 2x + 5 = 0
\]

To determine the number of solutions, we can use the discriminant (\(D\)) from the quadratic formula \(D = b^2 - 4ac\). In our equation \(x^2 - 2x + 5 = 0\):

- \(a = 1\)
- \(b = -2\)
- \(c = 5\)

Calculating the discriminant:

\[
D = (-2)^2 - 4(1)(5) = 4 - 20 = -16
\]

Since the discriminant is negative (\(D < 0\)), the quadratic has no real solutions.

Thus, the system of the linear and quadratic equations has **zero solutions**.

Final answer: This system has **0** solution(s).