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Section8.5Complex Solutions to Quadratic Equations

Figure8.5.1Alternative Video Lesson

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

Solution

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

Solution

\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.

Solution

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.

Solution

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.

Solution

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.

Solution

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.

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