Examples of quadratic equations demonstrating the method of completing the square.

Problem: Completing the Square

We are tasked with solving three quadratic equations using the method of completing the square. The equations are:

1. \( x^2 - 6x - 9 = 0 \)
2. \( 3x^2 - 5x - 7 = 0 \)
3. \( 7x^2 + 5x - 8 = 0 \)

Let's solve each equation step by step.

---

#### Equation 1: \( x^2 - 6x - 9 = 0 \)

1. Rearrange the equation:
\[
x^2 - 6x = 9
\]

2. Complete the square:
- Take the coefficient of \( x \), which is \(-6\).
- Divide it by 2: \(\frac{-6}{2} = -3\).
- Square it: \((-3)^2 = 9\).
- Add and subtract this square inside the equation:
\[
x^2 - 6x + 9 - 9 = 9
\]
\[
(x - 3)^2 - 9 = 9
\]

3. Simplify:
\[
(x - 3)^2 = 18
\]

4. Solve for \( x \):
\[
x - 3 = \pm \sqrt{18}
\]
\[
x - 3 = \pm 3\sqrt{2}
\]
\[
x = 3 \pm 3\sqrt{2}
\]

5. Final solution:
\[
x = 3 + 3\sqrt{2} \quad \text{or} \quad x = 3 - 3\sqrt{2}
\]

---

#### Equation 2: \( 3x^2 - 5x - 7 = 0 \)

1. Divide the entire equation by 3 to simplify:
\[
x^2 - \frac{5}{3}x - \frac{7}{3} = 0
\]

2. Rearrange the equation:
\[
x^2 - \frac{5}{3}x = \frac{7}{3}
\]

3. Complete the square:
- Take the coefficient of \( x \), which is \(-\frac{5}{3}\).
- Divide it by 2: \(\frac{-\frac{5}{3}}{2} = -\frac{5}{6}\).
- Square it: \(\left(-\frac{5}{6}\right)^2 = \frac{25}{36}\).
- Add and subtract this square inside the equation:
\[
x^2 - \frac{5}{3}x + \frac{25}{36} - \frac{25}{36} = \frac{7}{3}
\]
\[
\left(x - \frac{5}{6}\right)^2 - \frac{25}{36} = \frac{7}{3}
\]

4. Simplify:
- Convert \(\frac{7}{3}\) to a fraction with a denominator of 36:
\[
\frac{7}{3} = \frac{7 \cdot 12}{3 \cdot 12} = \frac{84}{36}
\]
- Substitute back:
\[
\left(x - \frac{5}{6}\right)^2 - \frac{25}{36} = \frac{84}{36}
\]
\[
\left(x - \frac{5}{6}\right)^2 = \frac{84}{36} + \frac{25}{36}
\]
\[
\left(x - \frac{5}{6}\right)^2 = \frac{109}{36}
\]

5. Solve for \( x \):
\[
x - \frac{5}{6} = \pm \sqrt{\frac{109}{36}}
\]
\[
x - \frac{5}{6} = \pm \frac{\sqrt{109}}{6}
\]
\[
x = \frac{5}{6} \pm \frac{\sqrt{109}}{6}
\]
\[
x = \frac{5 \pm \sqrt{109}}{6}
\]

6. Final solution:
\[
x = \frac{5 + \sqrt{109}}{6} \quad \text{or} \quad x = \frac{5 - \sqrt{109}}{6}
\]

---

#### Equation 3: \( 7x^2 + 5x - 8 = 0 \)

1. Divide the entire equation by 7 to simplify:
\[
x^2 + \frac{5}{7}x - \frac{8}{7} = 0
\]

2. Rearrange the equation:
\[
x^2 + \frac{5}{7}x = \frac{8}{7}
\]

3. Complete the square:
- Take the coefficient of \( x \), which is \(\frac{5}{7}\).
- Divide it by 2: \(\frac{\frac{5}{7}}{2} = \frac{5}{14}\).
- Square it: \(\left(\frac{5}{14}\right)^2 = \frac{25}{196}\).
- Add and subtract this square inside the equation:
\[
x^2 + \frac{5}{7}x + \frac{25}{196} - \frac{25}{196} = \frac{8}{7}
\]
\[
\left(x + \frac{5}{14}\right)^2 - \frac{25}{196} = \frac{8}{7}
\]

4. Simplify:
- Convert \(\frac{8}{7}\) to a fraction with a denominator of 196:
\[
\frac{8}{7} = \frac{8 \cdot 28}{7 \cdot 28} = \frac{224}{196}
\]
- Substitute back:
\[
\left(x + \frac{5}{14}\right)^2 - \frac{25}{196} = \frac{224}{196}
\]
\[
\left(x + \frac{5}{14}\right)^2 = \frac{224}{196} + \frac{25}{196}
\]
\[
\left(x + \frac{5}{14}\right)^2 = \frac{249}{196}
\]

5. Solve for \( x \):
\[
x + \frac{5}{14} = \pm \sqrt{\frac{249}{196}}
\]
\[
x + \frac{5}{14} = \pm \frac{\sqrt{249}}{14}
\]
\[
x = -\frac{5}{14} \pm \frac{\sqrt{249}}{14}
\]
\[
x = \frac{-5 \pm \sqrt{249}}{14}
\]

6. Final solution:
\[
x = \frac{-5 + \sqrt{249}}{14} \quad \text{or} \quad x = \frac{-5 - \sqrt{249}}{14}
\]

---

Final Answers:

\[
\boxed{
\begin{aligned}
1. & \quad x = 3 + 3\sqrt{2} \quad \text{or} \quad x = 3 - 3\sqrt{2} \\
2. & \quad x = \frac{5 + \sqrt{109}}{6} \quad \text{or} \quad x = \frac{5 - \sqrt{109}}{6} \\
3. & \quad x = \frac{-5 + \sqrt{249}}{14} \quad \text{or} \quad x = \frac{-5 - \sqrt{249}}{14}
\end{aligned}
}
\]