Dividing Polynomials Lesson Plans & Worksheets :: 25 - 48 - Free Printable

Problem Analysis:

The task involves solving a series of quadratic equations using the quadratic formula. The quadratic formula is given by:

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

where \(a\), \(b\), and \(c\) are coefficients from the standard form of a quadratic equation:

\[
ax^2 + bx + c = 0
\]

We will solve each part of the problem step by step.

---

Part 1: Solve \(3x^2 - 5x - 2 = 0\)

#### Step 1: Identify coefficients
From the equation \(3x^2 - 5x - 2 = 0\):
- \(a = 3\)
- \(b = -5\)
- \(c = -2\)

#### Step 2: Apply the quadratic formula
Substitute \(a\), \(b\), and \(c\) into the quadratic formula:

\[
x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(-2)}}{2(3)}
\]

Simplify inside the formula:
- \(-(-5) = 5\)
- \((-5)^2 = 25\)
- \(4(3)(-2) = -24\)
- \(2(3) = 6\)

So the formula becomes:

\[
x = \frac{5 \pm \sqrt{25 - (-24)}}{6}
\]

Simplify further:
\[
x = \frac{5 \pm \sqrt{25 + 24}}{6}
\]
\[
x = \frac{5 \pm \sqrt{49}}{6}
\]

Since \(\sqrt{49} = 7\):

\[
x = \frac{5 \pm 7}{6}
\]

#### Step 3: Solve for the two roots
- For the "+" case:
\[
x = \frac{5 + 7}{6} = \frac{12}{6} = 2
\]

- For the "-" case:
\[
x = \frac{5 - 7}{6} = \frac{-2}{6} = -\frac{1}{3}
\]

#### Final Answer for Part 1:
\[
x = 2 \quad \text{or} \quad x = -\frac{1}{3}
\]

---

Part 2: Solve \(2x^2 + 7x + 3 = 0\)

#### Step 1: Identify coefficients
From the equation \(2x^2 + 7x + 3 = 0\):
- \(a = 2\)
- \(b = 7\)
- \(c = 3\)

#### Step 2: Apply the quadratic formula
Substitute \(a\), \(b\), and \(c\) into the quadratic formula:

\[
x = \frac{-7 \pm \sqrt{7^2 - 4(2)(3)}}{2(2)}
\]

Simplify inside the formula:
- \(7^2 = 49\)
- \(4(2)(3) = 24\)
- \(2(2) = 4\)

So the formula becomes:

\[
x = \frac{-7 \pm \sqrt{49 - 24}}{4}
\]

Simplify further:
\[
x = \frac{-7 \pm \sqrt{25}}{4}
\]

Since \(\sqrt{25} = 5\):

\[
x = \frac{-7 \pm 5}{4}
\]

#### Step 3: Solve for the two roots
- For the "+" case:
\[
x = \frac{-7 + 5}{4} = \frac{-2}{4} = -\frac{1}{2}
\]

- For the "-" case:
\[
x = \frac{-7 - 5}{4} = \frac{-12}{4} = -3
\]

#### Final Answer for Part 2:
\[
x = -\frac{1}{2} \quad \text{or} \quad x = -3
\]

---

Part 3: Solve \(x^2 - 8x + 15 = 0\)

#### Step 1: Identify coefficients
From the equation \(x^2 - 8x + 15 = 0\):
- \(a = 1\)
- \(b = -8\)
- \(c = 15\)

#### Step 2: Apply the quadratic formula
Substitute \(a\), \(b\), and \(c\) into the quadratic formula:

\[
x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(15)}}{2(1)}
\]

Simplify inside the formula:
- \(-(-8) = 8\)
- \((-8)^2 = 64\)
- \(4(1)(15) = 60\)
- \(2(1) = 2\)

So the formula becomes:

\[
x = \frac{8 \pm \sqrt{64 - 60}}{2}
\]

Simplify further:
\[
x = \frac{8 \pm \sqrt{4}}{2}
\]

Since \(\sqrt{4} = 2\):

\[
x = \frac{8 \pm 2}{2}
\]

#### Step 3: Solve for the two roots
- For the "+" case:
\[
x = \frac{8 + 2}{2} = \frac{10}{2} = 5
\]

- For the "-" case:
\[
x = \frac{8 - 2}{2} = \frac{6}{2} = 3
\]

#### Final Answer for Part 3:
\[
x = 5 \quad \text{or} \quad x = 3
\]

---

Part 4: Solve \(x^2 + 4x + 4 = 0\)

#### Step 1: Identify coefficients
From the equation \(x^2 + 4x + 4 = 0\):
- \(a = 1\)
- \(b = 4\)
- \(c = 4\)

#### Step 2: Apply the quadratic formula
Substitute \(a\), \(b\), and \(c\) into the quadratic formula:

\[
x = \frac{-4 \pm \sqrt{4^2 - 4(1)(4)}}{2(1)}
\]

Simplify inside the formula:
- \(4^2 = 16\)
- \(4(1)(4) = 16\)
- \(2(1) = 2\)

So the formula becomes:

\[
x = \frac{-4 \pm \sqrt{16 - 16}}{2}
\]

Simplify further:
\[
x = \frac{-4 \pm \sqrt{0}}{2}
\]

Since \(\sqrt{0} = 0\):

\[
x = \frac{-4 \pm 0}{2}
\]

#### Step 3: Solve for the root
- For both cases (\(+\) and \(-\)):
\[
x = \frac{-4}{2} = -2
\]

#### Final Answer for Part 4:
\[
x = -2
\]

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Final Answers:

1. \(x = 2 \quad \text{or} \quad x = -\frac{1}{3}\)
2. \(x = -\frac{1}{2} \quad \text{or} \quad x = -3\)
3. \(x = 5 \quad \text{or} \quad x = 3\)
4. \(x = -2\)

\[
\boxed{x = 2 \text{ or } x = -\frac{1}{3}, \quad x = -\frac{1}{2} \text{ or } x = -3, \quad x = 5 \text{ or } x = 3, \quad x = -2}
\]