Solve for the following equations using the quadratic formula. Show your complete solution.

1. x² + 7x + 12 = 0
2. 2x² + 3x - 2 = 0
3. x² - 6x + 9 = 0

To solve each of these quadratic equations using the quadratic formula, we follow the formula:

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

where \(a\), \(b\), and \(c\) are the coefficients from the quadratic equation \(ax^2 + bx + c = 0\).

### 1. Solve \(x^2 + 7x + 12 = 0\)

- **Step 1: Identify the coefficients:**
  - \(a = 1\), \(b = 7\), \(c = 12\)

- **Step 2: Apply the quadratic formula:**
  \[
  x = \frac{-7 \pm \sqrt{7^2 - 4(1)(12)}}{2(1)}
  \]
  \[
  x = \frac{-7 \pm \sqrt{49 - 48}}{2}
  \]
  \[
  x = \frac{-7 \pm \sqrt{1}}{2}
  \]
  \[
  x = \frac{-7 \pm 1}{2}
  \]

- **Step 3: Solve for both possible values of \(x\):**
  - \(x_1 = \frac{-7 + 1}{2} = \frac{-6}{2} = -3\)
  - \(x_2 = \frac{-7 - 1}{2} = \frac{-8}{2} = -4\)

- **Solution:**
  \[
  x = -3 \quad \text{or} \quad x = -4
  \]

---

### 2. Solve \(2x^2 + 3x - 2 = 0\)

- **Step 1: Identify the coefficients:**
  - \(a = 2\), \(b = 3\), \(c = -2\)

- **Step 2: Apply the quadratic formula:**
  \[
  x = \frac{-3 \pm \sqrt{3^2 - 4(2)(-2)}}{2(2)}
  \]
  \[
  x = \frac{-3 \pm \sqrt{9 + 16}}{4}
  \]
  \[
  x = \frac{-3 \pm \sqrt{25}}{4}
  \]
  \[
  x = \frac{-3 \pm 5}{4}
  \]

- **Step 3: Solve for both possible values of \(x\):**
  - \(x_1 = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{1}{2}\)
  - \(x_2 = \frac{-3 - 5}{4} = \frac{-8}{4} = -2\)

- **Solution:**
  \[
  x = \frac{1}{2} \quad \text{or} \quad x = -2
  \]

---

### 3. Solve \(x^2 - 6x + 9 = 0\)

- **Step 1: Identify the coefficients:**
  - \(a = 1\), \(b = -6\), \(c = 9\)

- **Step 2: Apply the quadratic formula:**
  \[
  x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(9)}}{2(1)}
  \]
  \[
  x = \frac{6 \pm \sqrt{36 - 36}}{2}
  \]
  \[
  x = \frac{6 \pm \sqrt{0}}{2}
  \]
  \[
  x = \frac{6 \pm 0}{2}
  \]

- **Step 3: Solve for the single value of \(x\):**
  \[
  x = \frac{6}{2} = 3
  \]

- **Solution:**
  \[
  x = 3 \quad (\text{a repeated root})
  \]

---

### Summary of Solutions:
1. \(x = -3\) or \(x = -4\)
2. \(x = \frac{1}{2}\) or \(x = -2\)
3. \(x = 3\) (double root)

Would you like further clarification or details on any of the steps?

Here are 5 related questions for you:
1. How do we determine when a quadratic has real, repeated, or complex roots?
2. Can the quadratic formula be used for any type of quadratic equation?
3. How does the discriminant in the quadratic formula affect the number of solutions?
4. What happens when the value of \(b^2 - 4ac\) is negative?
5. Can you derive the quadratic formula from the process of completing the square?

**Tip:** The discriminant \(b^2 - 4ac\) helps to quickly determine the nature of the roots: positive (real and distinct), zero (real and repeated), or negative (complex).

Learn how to solve quadratic equations x² + 7x + 12, 2x² + 3x - 2, and x² - 6x + 9 using the quadratic formula. This guide provides detailed step-by-step solutions, covering key algebraic concepts such as roots of equations and discriminant analysis. Suitable for students in Grades 9-10.

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