Question

3x^2 + 8x + 4 9th grade algebra

Answer 1

To analyze and work with the quadratic expression \(3x^2 + 8x + 4\), we can perform several tasks commonly assigned in a 9th-grade algebra course, such as factoring, finding the vertex, completing the square, or determining the roots using the quadratic formula. Here’s a brief overview of these steps:

### 1. **Factoring**:
To factor the quadratic expression \(3x^2 + 8x + 4\), we look for two numbers that multiply to \(3 \times 4 = 12\) (the product of the coefficient of \(x^2\) and the constant term) and add to \(8\) (the coefficient of \(x\)).

The factors of 12 that work are \(6\) and \(2\), so we can rewrite \(8x\) as \(6x + 2x\):

\[
3x^2 + 6x + 2x + 4
\]

Next, we group the terms:

\[
(3x^2 + 6x) + (2x + 4)
\]

Factoring out the common factors from each group gives us:

\[
3x(x + 2) + 2(x + 2)
\]

We can now factor by grouping:

\[
(3x + 2)(x + 2)
\]

### 2. **Finding the roots**:
We can find the roots (values of \(x\) that make the expression equal to zero) by setting the factored form to zero:

\[
(3x + 2)(x + 2) = 0
\]

This gives us two equations to solve:

1. \(3x + 2 = 0\) → \(3x = -2\) → \(x = -\frac{2}{3}\)
2. \(x + 2 = 0\) → \(x = -2\)

### 3. **Using the quadratic formula**:
The quadratic formula can also be used to find the roots:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For \(3x^2 + 8x + 4\):
- \(a = 3\)
- \(b = 8\)
- \(c = 4\)

Calculating the discriminant:

\[
b^2 - 4ac = 8^2 - 4(3)(4) = 64 - 48 = 16
\]

Now plug into the quadratic formula:

\[
x = \frac{-8 \pm \sqrt{16}}{2 \cdot 3} = \frac{-8 \pm 4}{6}
\]

This gives us two solutions:

1. \(x = \frac{-8 + 4}{6} = \frac{-4}{6} = -\frac{2}{3}\)
2. \(x = \frac{-8 - 4}{6} = \frac{-12}{6} = -2\)

### 4. **Finding the vertex**:
The vertex of a parabola given by the equation \(ax^2 + bx + c\) can be found using the formula:

\[
x = -\frac{b}{2a}
\]

Substituting in our values:
\[
x = -\frac{8}{2 \cdot 3} = -\frac{8}{6} = -\frac{4}{3}
\]

To find the \(y\)-coordinate of the vertex, substitute \(x = -\frac{4}{3}\) back into the original equation:

\[
y = 3\left(-\frac{4}{3}\right)^2 + 8\left(-\frac{4}{3}\right) + 4 = 3\left(\frac{16}{9}\right) - \frac{32}{3} + 4
\]

Calculating:
\[
y = \frac{48}{9} - \frac{96}{9} + \frac{36}{9} = \frac{48 - 96 + 36}{9} = \frac{-12}{9} = -\frac{4}{3}
\]

So, the vertex is at \(\left(-\frac{4}{3}, -\frac{4}{3}\right)\).

### Summary:
- The roots are \(x = -\frac{2}{3}\) and \(x = -2\).
- The expression factors to \((3x + 2)(x + 2)\).
- The vertex is at \(\left(-\frac{4}{3}, -\frac{4}{3}\right)\).

Feel free to ask if you have more specific questions or need further explanations about any of these steps!