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find the roots of the quadratic equation by completing the square

To solve these quadratic equations by completing the square, we'll go through each step by step. Here are the quadratic equations:

1. \(x^2 + 6x + 5 = 0\)
2. \(x^2 + 4x - 6 = 0\)
3. \(x^2 - 8x - 20 = 0\)
4. \(x^2 - 6x - 10 = 0\)

### 1. \(x^2 + 6x + 5 = 0\)
- Move the constant term to the other side:
  \[
  x^2 + 6x = -5
  \]
- Take half of the coefficient of \(x\), square it, and add it to both sides:
  \[
  \left(\frac{6}{2}\right)^2 = 9
  \]
  \[
  x^2 + 6x + 9 = -5 + 9
  \]
- Rewrite the left side as a perfect square:
  \[
  (x + 3)^2 = 4
  \]
- Take the square root of both sides:
  \[
  x + 3 = \pm 2
  \]
- Solve for \(x\):
  \[
  x = -3 \pm 2
  \]
  So, \(x = -1\) or \(x = -5\).

### 2. \(x^2 + 4x - 6 = 0\)
- Move the constant term to the other side:
  \[
  x^2 + 4x = 6
  \]
- Take half of the coefficient of \(x\), square it, and add it to both sides:
  \[
  \left(\frac{4}{2}\right)^2 = 4
  \]
  \[
  x^2 + 4x + 4 = 6 + 4
  \]
- Rewrite the left side as a perfect square:
  \[
  (x + 2)^2 = 10
  \]
- Take the square root of both sides:
  \[
  x + 2 = \pm \sqrt{10}
  \]
- Solve for \(x\):
  \[
  x = -2 \pm \sqrt{10}
  \]
  So, \(x = -2 + \sqrt{10}\) or \(x = -2 - \sqrt{10}\).

### 3. \(x^2 - 8x - 20 = 0\)
- Move the constant term to the other side:
  \[
  x^2 - 8x = 20
  \]
- Take half of the coefficient of \(x\), square it, and add it to both sides:
  \[
  \left(\frac{-8}{2}\right)^2 = 16
  \]
  \[
  x^2 - 8x + 16 = 20 + 16
  \]
- Rewrite the left side as a perfect square:
  \[
  (x - 4)^2 = 36
  \]
- Take the square root of both sides:
  \[
  x - 4 = \pm 6
  \]
- Solve for \(x\):
  \[
  x = 4 \pm 6
  \]
  So, \(x = 10\) or \(x = -2\).

### 4. \(x^2 - 6x - 10 = 0\)
- Move the constant term to the other side:
  \[
  x^2 - 6x = 10
  \]
- Take half of the coefficient of \(x\), square it, and add it to both sides:
  \[
  \left(\frac{-6}{2}\right)^2 = 9
  \]
  \[
  x^2 - 6x + 9 = 10 + 9
  \]
- Rewrite the left side as a perfect square:
  \[
  (x - 3)^2 = 19
  \]
- Take the square root of both sides:
  \[
  x - 3 = \pm \sqrt{19}
  \]
- Solve for \(x\):
  \[
  x = 3 \pm \sqrt{19}
  \]
  So, \(x = 3 + \sqrt{19}\) or \(x = 3 - \sqrt{19}\).

### Summary of Solutions:
1. \(x = -1\) or \(x = -5\)
2. \(x = -2 + \sqrt{10}\) or \(x = -2 - \sqrt{10}\)
3. \(x = 10\) or \(x = -2\)
4. \(x = 3 + \sqrt{19}\) or \(x = 3 - \sqrt{19}\)

Would you like further details on any of these solutions or have any other questions?

Here are five related questions for further practice:
1. What is the general formula for completing the square for any quadratic equation \(ax^2 + bx + c = 0\)?
2. Can you apply completing the square to solve the quadratic equation \(2x^2 + 8x - 10 = 0\)?
3. How does the method of completing the square relate to the quadratic formula?
4. What are the conditions under which completing the square might be more advantageous than using the quadratic formula?
5. How can completing the square be used to derive the vertex form of a quadratic function?

**Tip:** Completing the square is also useful for converting quadratic functions into vertex form, which makes it easier to find the vertex and analyze the graph of the function.

Learn how to solve quadratic equations by completing the square with a detailed step-by-step guide. This tutorial covers quadratic equations such as x^2 + 6x + 5 = 0, x^2 + 4x - 6 = 0, x^2 - 8x - 20 = 0, and x^2 - 6x - 10 = 0. Understand the process of completing the square and find the roots of quadratic equations effectively.

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