Question

what is an example of an equation that could be solved by completing the square but not factoring? what is an example of one that could be solved more easily by factoring?

Answer 1

### Example of an Equation that can be Solved by Completing the Square but not Factoring:

Consider the quadratic equation:

\[
x^2 + 4x + 5 = 0
\]

This equation does not factor neatly into rational roots since the discriminant \(b^2 - 4ac = 4^2 - 4 \cdot 1 \cdot 5 = 16 - 20 = -4\) is negative. Thus, it does not have real solutions.

To solve this by completing the square:

1. Move the constant to the other side:
\[
x^2 + 4x = -5
\]

2. Complete the square:
\[
x^2 + 4x + 4 = -5 + 4
\]
\[
(x + 2)^2 = -1
\]

3. Take the square root of both sides:
\[
x + 2 = \pm i
\]

4. Solve for \(x\):
\[
x = -2 \pm i
\]

Thus, the solutions are \(x = -2 + i\) and \(x = -2 - i\).

### Example of an Equation that can be Solved More Easily by Factoring:

Consider the quadratic equation:

\[
x^2 - 5x + 6 = 0
\]

This equation factors easily:

1. We look for two numbers that multiply to \(6\) and add to \(-5\). The numbers \(-2\) and \(-3\) work.
2. Factor the equation:
\[
(x - 2)(x - 3) = 0
\]

3. Set each factor to zero:
\[
x - 2 = 0 \quad \Rightarrow \quad x = 2
\]
\[
x - 3 = 0 \quad \Rightarrow \quad x = 3
\]

Thus, the solutions are \(x = 2\) and \(x = 3\).

In summary:
- **Completing the square** is used for \(x^2 + 4x + 5 = 0\) without real solutions.
- **Factoring** is used for \(x^2 - 5x + 6 = 0\), which simplifies easily.