Question
Solve the system of equations.
y=14(x−1)2−3
y=−12x−12
Use the image to answer the question.
The coordinate grid has x and y axes labeled from negative 5 to 5. A U-shaped curve opens upward.
Enter the solution with the least x-value first.
(1 point)
The solutions are (
,
) and (
,
).
y=14(x−1)2−3
y=−12x−12
Use the image to answer the question.
The coordinate grid has x and y axes labeled from negative 5 to 5. A U-shaped curve opens upward.
Enter the solution with the least x-value first.
(1 point)
The solutions are (
,
) and (
,
).
Answers
GPT-4o mini
To solve the system of equations given:
1. \( y = 14(x - 1)^2 - 3 \)
2. \( y = -12x - 12 \)
We will set these two equations equal to each other because they both equal \( y \):
\[
14(x - 1)^2 - 3 = -12x - 12
\]
Next, we can simplify and solve for \( x \).
1. Start by expanding \( 14(x - 1)^2 \):
\[
14(x - 1)^2 = 14(x^2 - 2x + 1) = 14x^2 - 28x + 14
\]
So we rewrite the equation:
\[
14x^2 - 28x + 14 - 3 = -12x - 12
\]
This simplifies to:
\[
14x^2 - 28x + 11 = -12x - 12
\]
Next, move all terms to one side of the equation:
\[
14x^2 - 28x + 12x + 11 + 12 = 0
\]
Combining like terms:
\[
14x^2 - 16x + 23 = 0
\]
This is a quadratic equation in standard form \( ax^2 + bx + c = 0 \).
To solve for \( x \), we will use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 14 \), \( b = -16 \), and \( c = 23 \).
Calculate the discriminant:
\[
b^2 - 4ac = (-16)^2 - 4(14)(23) = 256 - 1288 = -1032
\]
Since the discriminant is negative, it indicates that there are no real solutions for this system. Thus, there are no intersection points between the two equations.
Therefore, we conclude:
The solutions are (no real solutions).
1. \( y = 14(x - 1)^2 - 3 \)
2. \( y = -12x - 12 \)
We will set these two equations equal to each other because they both equal \( y \):
\[
14(x - 1)^2 - 3 = -12x - 12
\]
Next, we can simplify and solve for \( x \).
1. Start by expanding \( 14(x - 1)^2 \):
\[
14(x - 1)^2 = 14(x^2 - 2x + 1) = 14x^2 - 28x + 14
\]
So we rewrite the equation:
\[
14x^2 - 28x + 14 - 3 = -12x - 12
\]
This simplifies to:
\[
14x^2 - 28x + 11 = -12x - 12
\]
Next, move all terms to one side of the equation:
\[
14x^2 - 28x + 12x + 11 + 12 = 0
\]
Combining like terms:
\[
14x^2 - 16x + 23 = 0
\]
This is a quadratic equation in standard form \( ax^2 + bx + c = 0 \).
To solve for \( x \), we will use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 14 \), \( b = -16 \), and \( c = 23 \).
Calculate the discriminant:
\[
b^2 - 4ac = (-16)^2 - 4(14)(23) = 256 - 1288 = -1032
\]
Since the discriminant is negative, it indicates that there are no real solutions for this system. Thus, there are no intersection points between the two equations.
Therefore, we conclude:
The solutions are (no real solutions).