To solve the system of equations:
[tex]\[
\begin{cases}
y = -4x \\
y = x^2 - 12
\end{cases}
\][/tex]
We need to find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy both equations simultaneously. Here's a detailed, step-by-step solution:
1. Set the equations equal to each other:
Since [tex]\( y \)[/tex] equals both [tex]\(-4x\)[/tex] and [tex]\( x^2 - 12 \)[/tex], we can set these expressions equal to each other:
[tex]\[
-4x = x^2 - 12
\][/tex]
2. Rearrange the equation to one side to form a quadratic equation:
Move all terms to one side of the equation to set it to zero:
[tex]\[
x^2 + 4x - 12 = 0
\][/tex]
3. Solve the quadratic equation:
To find the values of [tex]\( x \)[/tex], we can solve the quadratic equation using the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 4 \)[/tex], and [tex]\( c = -12 \)[/tex].
Calculate the discriminant:
[tex]\[
b^2 - 4ac = 4^2 - 4 \cdot 1 \cdot (-12) = 16 + 48 = 64
\][/tex]
Substitute back into the quadratic formula:
[tex]\[
x = \frac{-4 \pm \sqrt{64}}{2 \cdot 1} = \frac{-4 \pm 8}{2}
\][/tex]
So we have two solutions for [tex]\( x \)[/tex]:
[tex]\[
x = \frac{-4 + 8}{2} = \frac{4}{2} = 2
\][/tex]
[tex]\[
x = \frac{-4 - 8}{2} = \frac{-12}{2} = -6
\][/tex]
4. Find the corresponding [tex]\( y \)[/tex]-values:
Substitute each [tex]\( x \)[/tex]-value back into one of the original equations (let’s use [tex]\( y = -4x \)[/tex]) to find the corresponding [tex]\( y \)[/tex]-values.
For [tex]\( x = 2 \)[/tex]:
[tex]\[
y = -4(2) = -8
\][/tex]
For [tex]\( x = -6 \)[/tex]:
[tex]\[
y = -4(-6) = 24
\][/tex]
5. Write the solutions as ordered pairs:
The solutions to the system of equations are the points where the equations intersect. These points are:
[tex]\[
(-6, 24) \quad \text{and} \quad (2, -8)
\][/tex]
Therefore, the solution to the system of equations is:
A. [tex]\((-6, 24)\)[/tex] and [tex]\((2, -8)\)[/tex]
