Lesson 17

Completing the Square and Complex Solutions

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

17.1: Creating Quadratic Equations

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 \(6 - i \sqrt{5}\).