Question

Use the elimination method to solve the following linear-quadratic system:

y=12x^2−8x−2

y=−3x

(1 point)

Answer 1

is 1/4 negative or postive?

Answer 2

To solve the system of equations using the elimination method, we start with the given equations:

1. \( y = 12x^2 - 8x - 2 \) (Equation 1)
2. \( y = -3x \) (Equation 2)

### Step 1: Set the equations equal to each other

Since both equations are equal to \( y \), we can set them equal to each other:

\[
12x^2 - 8x - 2 = -3x
\]

### Step 2: Rearrange the equation

Now, we will move all terms to one side of the equation in order to set it to zero:

\[
12x^2 - 8x + 3x - 2 = 0
\]
\[
12x^2 - 5x - 2 = 0
\]

### Step 3: Solve the quadratic equation

We can use the quadratic formula to find the values of \( x \):

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 12 \), \( b = -5 \), and \( c = -2 \).

Calculating the discriminant:

\[
b^2 - 4ac = (-5)^2 - 4 \cdot 12 \cdot (-2)
\]
\[
= 25 + 96 = 121
\]

Now applying the quadratic formula:

\[
x = \frac{-(-5) \pm \sqrt{121}}{2 \cdot 12}
\]
\[
= \frac{5 \pm 11}{24}
\]

Calculating the two possible solutions for \( x \):

1. \( x = \frac{5 + 11}{24} = \frac{16}{24} = \frac{2}{3} \)
2. \( x = \frac{5 - 11}{24} = \frac{-6}{24} = -\frac{1}{4} \)

### Step 4: Find the corresponding \( y \) values

Now that we have our \( x \) values, we can substitute them back into one of the original equations to find the corresponding \( y \) values. We will use Equation 2 (\( y = -3x \)) for this step.

**For \( x = \frac{2}{3} \):**

\[
y = -3\left(\frac{2}{3}\right) = -2
\]
Thus, one solution is \( \left(\frac{2}{3}, -2\right) \).

**For \( x = -\frac{1}{4} \):**

\[
y = -3\left(-\frac{1}{4}\right) = \frac{3}{4}
\]
Thus, the other solution is \( \left(-\frac{1}{4}, \frac{3}{4}\right) \).

### Final Solutions

The solutions to the system of equations are:

1. \( \left(\frac{2}{3}, -2\right) \)
2. \( \left(-\frac{1}{4}, \frac{3}{4}\right) \)

Answer 3

The fraction \(-\frac{1}{4}\) is negative. The negative sign indicates that it is less than zero. In contrast, \(\frac{1}{4}\) (without the negative sign) is positive, as it is greater than zero.

So to clarify:
- \(-\frac{1}{4}\) is negative.
- \(\frac{1}{4}\) is positive.