Solve the system of equations.

Question

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 (
,
).

Answer 1

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).