## Section8.5Complex Solutions to Quadratic Equations

¶### Subsection8.5.1Imaginary 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**.

###### Definition8.5.2Imaginary Numbers

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

###### Example8.5.3

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

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

###### Example8.5.4

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

### Subsection8.5.2Solving Quadratic Equations with Imaginary Solutions

###### Example8.5.5

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

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 \(\{-7i,7i\}\text{.}\)

###### Example8.5.6

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

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 \(\left\{-5i\sqrt{3},5i\sqrt{3}\right\}\text{.}\)

### Subsection8.5.3Solving Quadratic Equations with Complex Solutions

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

###### Definition8.5.7Complex 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**. You can read more at en.wikipedia.org/wiki/Complex_number.

Here are some examples of equations that have complex solutions.

###### Example8.5.8

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

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

###### Example8.5.9

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

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 8.5.9, 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.

###### Remark8.5.10

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 8.5.9, 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*}### Subsection8.5.4Exercises

Simplifying Square Roots with Negative Radicands

###### 1

Simplify the radical and write it into a complex number.

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

###### 2

Simplify the radical and write it into a complex number.

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

###### 3

Simplify the radical and write it into a complex number.

\(\displaystyle{ \sqrt{-54} =}\)

###### 4

Simplify the radical and write it into a complex number.

\(\displaystyle{ \sqrt{-50} =}\)

###### 5

Simplify the radical and write it into a complex number.

\(\displaystyle{ \sqrt{-168} =}\)

###### 6

Simplify the radical and write it into a complex number.

\(\displaystyle{ \sqrt{-250} =}\)

Quadratic Equations with Imaginary and Complex Solutions

###### 7

Solve the quadratic equation. Solutions could be complex numbers.

\(t^2 = -49\)

###### 8

Solve the quadratic equation. Solutions could be complex numbers.

\(x^2 = -16\)

###### 9

Solve the quadratic equation. Solutions could be complex numbers.

\(6x^2 - 10 = -610\)

###### 10

Solve the quadratic equation. Solutions could be complex numbers.

\(-4x^2 - 10 = 186\)

###### 11

Solve the quadratic equation. Solutions could be complex numbers.

\({5y^{2}} + 4 = -6\)

###### 12

Solve the quadratic equation. Solutions could be complex numbers.

\({2y^{2}} - 2 = -8\)

###### 13

Solve the quadratic equation. Solutions could be complex numbers.

\(4r^2+10 = -242\)

###### 14

Solve the quadratic equation. Solutions could be complex numbers.

\(-10r^2 - 3 = 277\)

###### 15

Solve the quadratic equation. Solutions could be complex numbers.

\(-9(t+8)^2+10 = 739\)

###### 16

Solve the quadratic equation. Solutions could be complex numbers.

\(7(t - 3)^2+9 = -243\)

###### 17

Solve the quadratic equation. Solutions could be complex numbers.

\({x^{2}-2x+5} = 0\)

###### 18

Solve the quadratic equation. Solutions could be complex numbers.

\({x^{2}+6x+10} = 0\)

###### 19

Solve the quadratic equation. Solutions could be complex numbers.

\({x^{2}+10x+28} =0\)

###### 20

Solve the quadratic equation. Solutions could be complex numbers.

\({y^{2}+8y+19} =0\)