Question

Use the elimination method to determine whether the linear-quadratic system y=x2−4x+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 the number of solutions for the given system of equations using the elimination method, we will rewrite the equations and analyze them.

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

To find the intersection points, we can set the two equations equal to each other:
\[
x^2 - 4x + 10 = -2x + 5
\]

Now, let's rearrange the equation:
\[
x^2 - 4x + 2x + 10 - 5 = 0
\]
\[
x^2 - 2x + 5 = 0
\]

Now we will use the quadratic formula to solve for \(x\):
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \(a = 1\), \(b = -2\), and \(c = 5\).

Calculating the discriminant \(b^2 - 4ac\):
\[
(-2)^2 - 4(1)(5) = 4 - 20 = -16
\]

Since the discriminant is negative (-16), this indicates that there are no real solutions to the equation. This means that the quadratic and linear equations do not intersect.

Thus, the system has **0 solutions**.

The correct answer is:
**0**