# Lesson 17

Completing the Square and Complex Solutions

• Let’s find complex solutions to quadratic equations by completing the square.

Match each equation in standard form to its factored form and its solutions.

1. $$x^2 - 25 = 0$$
2. $$x^2 - 5 = 0$$
3. $$x^2 + 25 = 0$$
• $$(x - 5i)(x + 5i) = 0$$
• $$(x - 5)(x + 5) = 0$$
• $$(x - \sqrt{5})(x + \sqrt{5}) = 0$$
• $$\sqrt{5}$$$$\text- \sqrt{5}$$
• 5, -5
• $$5i$$, ​​​​$$\text- 5i$$

### 17.2: Sometimes the Solutions Aren't Real Numbers

What are the solutions to these equations?

1. $$(x - 5)^2 = 0$$
2. $$(x - 5)^2 = 1$$
3. $$(x - 5)^2 = \text- 1$$

### 17.3: Finding Complex Solutions

Solve these equations by completing the square.

1. $$x^2 - 8x + 13 = 0$$
2. $$x^2 - 8x + 19 = 0$$

For which values of $$a$$ does the equation $$x^2 - 8x + a = 0$$ have two real solutions? One real solution? No real solutions? Explain your reasoning.

### 17.4: Can You See the Solutions on a Graph?

1. How many real solutions does each equation have? How many non-real solutions?
1. $$x^2 - 8x + 13 = 0$$
2. $$x^2 - 8x + 16 = 0$$
3. $$x^2 - 8x + 19 = 0$$
2. How do the graphs of these functions help us answer the previous question?
1. $$f(x) = x^2 - 8x + 13$$
2. $$g(x) = x^2 - 8x + 16$$
3. $$h(x) = x^2 - 8x + 19$$

### Summary

Sometimes quadratic equations have real solutions, and sometimes they do not. Here is a quadratic equation with $$x^2$$ equal to a negative number (assume $$k$$ is positive):

$$\displaystyle x^2 = \text-k$$

This equation will have imaginary solutions $$i\sqrt{k}$$ and $$\text-i\sqrt{k}$$. By similar reasoning, an equation of the form:

$$\displaystyle (x- h)^2 = \text-k$$

will have non-real solutions if $$k$$ is positive. In this case, the solutions are $$h+ i\sqrt{k}$$ and $$h- i\sqrt{k}$$.

It isn’t always clear just by looking at a quadratic equation whether the solutions will be real or not. For example, look at this quadratic equation:

$$\displaystyle x^2 - 12x + 41 = 0$$

We can always complete the square to find out what the solutions will be:

\displaystyle \begin{align} x^2 - 12x + 36 + 5 &= 0 \\ (x - 6)^2 + 5 &= 0 \\ (x - 6)^2 &= \text- 5 \\ x-6 &= \pm i \sqrt{5} \\ x &= 6 \pm i \sqrt{5} \end{align}

This equation has non-real, complex solutions $$6 + i \sqrt{5}$$ and $$6 - i \sqrt{5}$$.