m^2+20m=-36 solve by completing the square

To solve the quadratic equation \( m^2 + 20m = -36 \) by completing the square, follow these steps:

1. **Move the constant term to the other side**:  
   \[
   m^2 + 20m + \textcolor{blue}{\square} = -36 + \textcolor{blue}{\square}
   \]

2. **Find the value needed to complete the square**:  
   Take half of the coefficient of \( m \), square it, and add this value to both sides of the equation.  
   \[
   \text{Half of 20 is } 10, \text{ and } 10^2 = 100.
   \]
   So, add 100 to both sides:
   \[
   m^2 + 20m + 100 = -36 + 100
   \]

3. **Simplify the equation**:  
   \[
   m^2 + 20m + 100 = 64
   \]

4. **Express the left side as a perfect square**:  
   \[
   (m + 10)^2 = 64
   \]

5. **Solve for \( m \) by taking the square root of both sides**:  
   \[
   m + 10 = \pm 8
   \]
   So, the two possible solutions are:
   \[
   m + 10 = 8 \quad \text{or} \quad m + 10 = -8
   \]

6. **Isolate \( m \)**:  
   For \( m + 10 = 8 \):
   \[
   m = 8 - 10 = -2
   \]
   For \( m + 10 = -8 \):
   \[
   m = -8 - 10 = -18
   \]

Thus, the solutions are \( m = -2 \) and \( m = -18 \).

Would you like more details or have any other questions?

Here are some related questions:

1. How do you complete the square for equations with a leading coefficient other than 1?
2. What are the advantages of completing the square versus using the quadratic formula?
3. How does completing the square relate to the vertex form of a quadratic equation?
4. Can completing the square be used to solve quadratic inequalities?
5. How do you derive the quadratic formula using the method of completing the square?

**Tip**: Completing the square is also useful in deriving the quadratic formula and in transforming quadratic functions into their vertex form.

m^2 + 20m = -36 solve by completing the square

This guide demonstrates how to solve the quadratic equation m^2 + 20m = -36 using the completing the square method. Follow the step-by-step process, covering key algebraic concepts and perfect square properties, suitable for Grades 8-10.

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