ORCCA Complex Solutions to Quadratic Equations

Section 9.4 Complex Solutions to Quadratic Equations

Subsection 9.4.1 Imaginary Numbers

Let's look at how to simplify a square root that has a negative radicand. Remember that \(\sqrt{16}=4\) because \(4^2=16\text{.}\) So what could \(\sqrt{-16}\) be equal to? There is no real number that we can square to get \(-16\text{,}\) because when you square a real number, the result is either positive or \(0\text{.}\) You might think about \(4\) and \(-4\text{,}\) but:

\begin{equation*} 4^2=16\text{ and }(-4)^2=16 \end{equation*}

so neither of those could be \(\sqrt{-16}\text{.}\) To handle this situation, mathematicians separate a factor of \(\sqrt{-1}\) and represent it with the letter \(i\text{,}\) which stands for imaginary unit.

Definition 9.4.1. Imaginary Numbers.

The imaginary unit, \(i\text{,}\) is defined by \(i=\sqrt{-1}\text{.}\) The imaginary unit ¹ satisfies the equation \(i^2=-1\text{.}\) A real number times \(i\text{,}\) such as \(4i\text{,}\) is called an imaginary number.

en.wikipedia.org/wiki/Imaginary_number

Now we can simplify square roots with negative radicands like \(\sqrt{-16}\text{.}\)

\begin{align*} \sqrt{-16}\amp=\sqrt{-1\cdot16}\\ \amp=\sqrt{-1}\cdot\sqrt{16}\\ \amp=i\cdot4\\ \amp=4i \end{align*}

Imaginary numbers are widely used in electrical engineering, physics, computer science and other fields. Let's look some more examples.

Example 9.4.2.

Simplify \(\sqrt{-2}\text{.}\)

Explanation

\begin{align*} \sqrt{-2}\amp=\sqrt{-1\cdot2}\\ \amp=\sqrt{-1}\cdot\sqrt{2}\\ \amp=i\sqrt{2} \end{align*}

We write the \(i\) first because it's difficult to tell the difference between \(\sqrt{2}i\) and \(\sqrt{2i}\text{.}\)

Example 9.4.3.

Simplify \(\sqrt{-72}\text{.}\)

Explanation

\begin{align*} \sqrt{-72}\amp=\sqrt{-1\cdot36\cdot2}\\ \amp=\sqrt{-1}\cdot\sqrt{36}\cdot\sqrt{2}\\ \amp=6i\sqrt{2} \end{align*}

Subsection 9.4.2 Solving Quadratic Equations with Imaginary Solutions

Example 9.4.4.

Solve for \(x\) in \(x^2+49=0\text{,}\) where \(x\) is an imaginary number.

Explanation

There is no \(x\) term so we will use the square root method.

\begin{align*} x^2+49\amp=0\\ x^2\amp=-49 \end{align*}

\begin{align*} x\amp=-\sqrt{-49} \amp\text{ or }\amp\amp x\amp=\sqrt{-49}\\ x\amp=-\sqrt{-1}\cdot\sqrt{49} \amp\text{ or }\amp\amp x\amp=\sqrt{-1}\cdot\sqrt{49}\\ x\amp=-7i\amp\text{ or }\amp\amp x\amp=7i \end{align*}

The solution set is \(\{-7i,7i\}\text{.}\)

Example 9.4.5.

Solve for \(p\) in \(p^2+75=0\text{,}\) where \(p\) is an imaginary number.

Explanation

There is no \(p\) term so we will use the square root method.

\begin{align*} p^2+75\amp=0\\ p^2\amp=-75 \end{align*}

\begin{align*} p\amp=-\sqrt{-75} \amp\text{ or }\amp\amp p\amp=\sqrt{-75}\\ p\amp=-\sqrt{-1}\cdot\sqrt{25}\cdot\sqrt{3} \amp\text{ or }\amp\amp p\amp=\sqrt{-1}\cdot\sqrt{25}\cdot\sqrt{3}\\ p\amp=-5i\sqrt{3}\amp\text{ or }\amp\amp p\amp=5i\sqrt{3} \end{align*}

The solution set is \(\left\{-5i\sqrt{3},5i\sqrt{3}\right\}\text{.}\)

Subsection 9.4.3 Solving Quadratic Equations with Complex Solutions

A complex number is a combination of a real number and an imaginary number, like \(3+2i\) or \(-4-8i\text{.}\)

Definition 9.4.6. Complex Number.

A complex number is a number that can be expressed in the form \(a + bi\text{,}\) where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit. In this expression, \(a\) is the real part and \(b\) (not \(bi\)) is the imaginary part of the complex number ².

en.wikipedia.org/wiki/Complex_number

Here are some examples of equations that have complex solutions.

Example 9.4.7.

Solve for \(m\) in \((m-1)^2+18=0\text{,}\) where \(m\) is a complex number.

Explanation

This equation has a squared expression so we will use the square root method.

\begin{align*} (m-1)^2+18\amp=0\\ (m-1)^2\amp=-18 \end{align*}

\begin{align*} m-1\amp=-\sqrt{-18} \amp\text{ or }\amp\amp m-1\amp=\sqrt{-18}\\ m-1\amp=-\sqrt{-1}\cdot\sqrt{9}\cdot\sqrt{2}\amp\text{ or }\amp\amp m-1\amp=\sqrt{-1}\cdot\sqrt{9}\cdot\sqrt{2}\\ m-1\amp=-3i\sqrt{2}\amp\text{ or }\amp\amp m-1\amp=3i\sqrt{2}\\ m\amp=1-3i\sqrt{2}\amp\text{ or }\amp\amp m\amp=1+3i\sqrt{2} \end{align*}

The solution set is \(\left\{1-3i\sqrt{2}, 1+3i\sqrt{2}\right\}\text{.}\)

Example 9.4.8.

Solve for \(y\) in \(y^2-4y+13=0\text{,}\) where \(y\) is a complex number.

Explanation

Note that there is a \(y\) term, but the left side does not factor. We will use the quadratic formula. We identify that \(a=1\text{,}\) \(b=-4\) and \(c=13\) and substitute them into the quadratic formula.

\begin{align*} y\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \amp=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(13)}}{2(1)}\\ \amp=\frac{4\pm\sqrt{16-52}}{2}\\ \amp=\frac{4\pm\sqrt{-36}}{2}\\ \amp=\frac{4\pm\sqrt{-1}\cdot\sqrt{36}}{2}\\ \amp=\frac{4\pm6i}{2}\\ \amp=2\pm 3i \end{align*}

The solution set is \(\{2- 3i, 2+ 3i\}\text{.}\)

Note that in Example 9.4.8, the expressions \(2+3i\) and \(2-3i\) are fully simplified. In the same way that the terms \(2\) and \(3x\) cannot be combined, the terms \(2\) and \(3i\) can not be combined.

Remark 9.4.9.

Each complex solution can be checked, just as every real solution can be checked. For example, to check the solution of \(2+3i\) from Example 9.4.8, we would replace \(y\) with \(2+3i\) and check that the two sides of the equation are equal. In doing so, we will need to use the fact that \(i^2=-1\text{.}\) This check is shown here:

\begin{align*} y^2-4y+13\amp=0\\ (\substitute{2+3i})^2-4(\substitute{2+3i})+13\amp\stackrel{?}{=}0\\ (2^2+2(3i)+2(3i)+(3i)^2)-4\cdot 2 -4\cdot (3i) +13 \amp\stackrel{?}{=}0\\ 4+6i+6i+9i^2-8-12i+13 \amp\stackrel{?}{=}0\\ 4+9(-1)-8+13 \amp\stackrel{?}{=}0\\ 4-9-8+13 \amp\stackrel{?}{=}0\\ 0\amp\stackrel{\checkmark}{=}0 \end{align*}

Exercises 9.4.4 Exercises

Simplifying Square Roots with Negative Radicands

1.

Simplify the radical and write it as a complex number using \(i\text{.}\)

\(\displaystyle{ \sqrt{-42} = }\)

2.

Simplify the radical and write it as a complex number using \(i\text{.}\)

\(\displaystyle{ \sqrt{-30} = }\)

3.

Simplify the radical and write it as a complex number using \(i\text{.}\)

\(\displaystyle{ \sqrt{-72} =}\)

4.

Simplify the radical and write it as a complex number using \(i\text{.}\)

\(\displaystyle{ \sqrt{-80} =}\)

5.

Simplify the radical and write it as a complex number using \(i\text{.}\)

\(\displaystyle{ \sqrt{-216} =}\)

6.

Simplify the radical and write it as a complex number using \(i\text{.}\)

\(\displaystyle{ \sqrt{-200} =}\)