## 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 1 en.wikipedia.org/wiki/Imaginary_number 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 $\sqrt{-2}\text{.}$

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 $\sqrt{2i}\text{.}$

###### Example8.5.4

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*}

### 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.

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 $\{-7i,7i\}\text{.}$

###### Example8.5.6

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

### 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 of the complex number 2 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.

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

###### Example8.5.9

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 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

###### 1

Simplify the radical and write it into a complex number.

$\displaystyle{ \sqrt{-105} = }$

###### 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.

$\displaystyle{ \sqrt{-72} =}$

###### 4

Simplify the radical and write it into a complex number.

$\displaystyle{ \sqrt{-56} =}$

###### 5

Simplify the radical and write it into a complex number.

$\displaystyle{ \sqrt{-126} =}$

###### 6

Simplify the radical and write it into a complex number.

$\displaystyle{ \sqrt{-200} =}$

###### 7

Solve the quadratic equation. Solutions could be complex numbers.

$y^2 = -16$

###### 8

Solve the quadratic equation. Solutions could be complex numbers.

$r^2 = -100$

###### 9

Solve the quadratic equation. Solutions could be complex numbers.

$-6r^2+2 = 296$

###### 10

Solve the quadratic equation. Solutions could be complex numbers.

$-3t^2+2 = 50$

###### 11

Solve the quadratic equation. Solutions could be complex numbers.

${2t^{2}} + 10 = 6$

###### 12

Solve the quadratic equation. Solutions could be complex numbers.

${2t^{2}} - 2 = -8$

###### 13

Solve the quadratic equation. Solutions could be complex numbers.

$-6x^2+4 = 124$

###### 14

Solve the quadratic equation. Solutions could be complex numbers.

$-6x^2 - 7 = 293$

###### 15

Solve the quadratic equation. Solutions could be complex numbers.

$9(y+4)^2+1 = -323$

###### 16

Solve the quadratic equation. Solutions could be complex numbers.

$6(y - 5)^2 - 7 = -61$

###### 17

Solve the quadratic equation. Solutions could be complex numbers.

${r^{2}+2r+5} = 0$

###### 18

Solve the quadratic equation. Solutions could be complex numbers.

${r^{2}+6r+10} = 0$

###### 19

Solve the quadratic equation. Solutions could be complex numbers.

${t^{2}+4t+9} =0$

###### 20

Solve the quadratic equation. Solutions could be complex numbers.

${t^{2}+4t+11} =0$