Question

Determine the solution(s) to the following system. y=−x2 y=12x−3 (1 point) Responses (1,−1) and (3,−9) left parenthesis 1 comma negative 1 right parenthesis and left parenthesis 3 comma negative 9 right parenthesis (2,−4) and (−1.5,−2.25) left parenthesis 2 comma negative 4 right parenthesis and left parenthesis negative 1.5 comma negative 2.25 right parenthesis (−2,−4) and (1.5,−2.25) left parenthesis negative 2 comma negative 4 right parenthesis and left parenthesis 1.5 comma negative 2.25 right parenthesis (1.5,−2.25)

Answer 1

To determine the solutions to the system of equations given by \( y = -x^2 \) and \( y = 12x - 3 \), we will set the two equations equal to each other and solve for \( x \):

1. Set the equations equal:
\[
-x^2 = 12x - 3
\]

2. Rearrange to form a standard quadratic equation:
\[
-x^2 - 12x + 3 = 0
\]
or multiplying through by -1:
\[
x^2 + 12x - 3 = 0
\]

3. Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 1 \), \( b = 12 \), and \( c = -3 \):
\[
b^2 - 4ac = 12^2 - 4 \cdot 1 \cdot (-3) = 144 + 12 = 156
\]

Thus, we can solve for \( x \):
\[
x = \frac{-12 \pm \sqrt{156}}{2}
\]
Simplifying \( \sqrt{156} \):
\[
\sqrt{156} = \sqrt{4 \cdot 39} = 2\sqrt{39}
\]

Therefore,
\[
x = \frac{-12 \pm 2\sqrt{39}}{2} = -6 \pm \sqrt{39}
\]

This gives us two \( x \) values:
\[
x_1 = -6 + \sqrt{39}, \quad x_2 = -6 - \sqrt{39}
\]

4. To find the corresponding \( y \) values, we can substitute these \( x \) values back into either of the original equations. We’ll use \( y = 12x - 3 \):
For \( x_1 \):
\[
y_1 = 12(-6 + \sqrt{39}) - 3 = -72 + 12\sqrt{39} - 3 = -75 + 12\sqrt{39}
\]
For \( x_2 \):
\[
y_2 = 12(-6 - \sqrt{39}) - 3 = -72 - 12\sqrt{39} - 3 = -75 - 12\sqrt{39}
\]

Therefore, the solutions to the system are:
\[
\left(-6 + \sqrt{39}, -75 + 12\sqrt{39}\right) \quad \text{and} \quad \left(-6 - \sqrt{39}, -75 - 12\sqrt{39}\right)
\]

None of the provided options exactly match those calculated solutions. If you provide further context or clarify the options, I could assist more accurately!