Question

Solve the linear-quadratic system

y=x2−4x−41
and

y=4


(1 point)
Responses

(5,4)
open paren 5 comma 4 close paren

(5, 4)
and (−9, 4)
(5, 4)
and (−9, 4)

(−5, 4)
and (9, 4)
(−5, 4)
and (9, 4)

(9, 4)

Answer 1

To solve the linear-quadratic system given by the equations:

1. \( y = x^2 - 4x - 41 \)
2. \( y = 4 \)

we will substitute the second equation into the first equation.

1. Set \( y = 4 \) in the quadratic equation:

\[
4 = x^2 - 4x - 41
\]

2. Rearranging gives:

\[
x^2 - 4x - 41 - 4 = 0
\]

\[
x^2 - 4x - 45 = 0
\]

3. Next, we can factor the quadratic:

\[
(x - 9)(x + 5) = 0
\]

This gives us the solutions for \( x \):

\[
x - 9 = 0 \quad \Rightarrow \quad x = 9
\]
\[
x + 5 = 0 \quad \Rightarrow \quad x = -5
\]

4. Now, we substitute these \( x \) values back to find corresponding \( y \):

When \( x = 9 \):

\[
y = 4
\]
So, one solution is \( (9, 4) \).

When \( x = -5 \):

\[
y = 4
\]
So, the other solution is \( (-5, 4) \).

Thus, the solutions to the system are:

\[
(9, 4) \quad \text{and} \quad (-5, 4)
\]

The correct response is:

**(−5, 4) and (9, 4)**