ORCCA Complex Solutions to Quadratic Equations

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Section 7.3 Complex Solutions to Quadratic Equations

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Objectives: PCC Course Content and Outcome Guide

MTH 65 CCOG

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Figure 7.3.1. Alternative Video Lesson

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Subsection 7.3.1 Imaginary Numbers

Let's take a closer look at a square root with a negative radicand. Remember that $\sqrt{16}=4$ because $4\cdot4=16\text{.}$ So what about $\sqrt{-16}\text{?}$ 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:

4 \cdot 4 = 16 and (- 4) (- 4) = 16

$\begin{equation*} 4\cdot4=16\text{ and }(-4)(-4)=16 \end{equation*}$

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 $i\text{.}$

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Definition 7.3.2. 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.

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

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

permalinkImaginary numbers are used in electrical engineering, physics, computer science, and advanced mathematics. Let's look some more examples.

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Example 7.3.3.

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 $i$ in front of the radical because it can be easy to mix up $\sqrt{2}i$ and $\sqrt{2i}\text{,}$ if you don't draw the radical very carefully.

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Example 7.3.4.

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*}

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Subsection 7.3.2 Solving Quadratic Equations with Imaginary Solutions

permalinkBack in Example 7.1.9, we examined an equation that had no real solution. Let's revisit that example now that we are aware of imaginary numbers.

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Example 7.3.5.

Solve for $x$ in $x^2+49=0\text{,}$ where $x$ might not be a real 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\\ x\amp=\pm\sqrt{-49}\\ x\amp=\pm\sqrt{-1\cdot49}\\ x\amp=\pm\sqrt{-1}\cdot\sqrt{49}\\ x\amp=\pm i\cdot7 \end{align*}

\begin{align*} x\amp=-7i\amp\text{ or }\amp\amp x\amp=7i \end{align*}

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

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Example 7.3.6.

Solve for $p$ in $p^2+75=0\text{,}$ where $p$ might not be a real 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\\ p\amp=\pm\sqrt{-75}\\ p\amp=\pm\sqrt{-1\cdot25\cdot3}\\ p\amp=\pm\sqrt{-1}\cdot\sqrt{25}\cdot\sqrt{3}\\ p\amp=\pm i\cdot5\sqrt{3} \end{align*}

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

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

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Subsection 7.3.3 Solving Quadratic Equations with Complex Solutions

Sometimes we need to work with a sum of a real number and an imaginary number, like $3+2i$ or $-4-8i\text{.}$ These combinations are called “complex numbers”.

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Definition 7.3.7. 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 ²en.wikipedia.org/wiki/Complex_number.

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Example 7.3.8.

In an advanced math course, you might study the relationship between a lynx polulation (or any generic predator) and a hare population (or any generic prey) as time passes. For example, if the predator population is high, they will eat many prey. But then the prey population will become low, so the predators will go hungry and have fewer offspring. With time, the predator population will decline, and that will lead to a rebound in the prey population. Then prey will be plentiful, and the preadator population will rebound, and the whole situaiton starts over. This cycle may take years or even decades to play out.

Strange as it may seem, to understand this phenomenon mathematically, you will need to solve equations similar to:

$\begin{equation*} (1-t)(3-t)+10=0 \end{equation*}$

Let's practice solving this equation.

$\begin{align*} (1-t)(3-t)+10\amp=0\\ 3-t-3t+t^2+10\amp=0\\ t^2-4t+13\amp=0 \end{align*}$

We can try the quadratic formula.

$\begin{align*} t\amp=\frac{4\pm\sqrt{(-4)^2-4(1)(13)}}{2(1)}\\ \amp=\frac{4\pm\sqrt{16-52}}{2}\\ \amp=\frac{4\pm\sqrt{-36}}{2}\\ \amp=\frac{4\pm\sqrt{-1}\cdot\sqrt{36}}{2}\\ \amp=\frac{4\pm i\cdot6}{2}\\ \amp=2\pm3i \end{align*}$

These two solutions, $2-3i$ and $2+3i$ have implications for how fast the predator and prey populations rise and fall over time, but an explanation is beyond the scope of basic algebra.

permalinkHere are some more examples of equations that have complex number solutions.

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Example 7.3.9.

Solve for $m$ in $(m-1)^2+18=0\text{,}$ where $m$ might not be a real 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\\ m-1\amp=\pm\sqrt{-18}\\ m-1\amp=\pm\sqrt{-1\cdot9\cdot2}\\ m-1\amp=\pm\sqrt{-1}\cdot\sqrt{9}\cdot\sqrt{2}\\ m-1\amp=\pm i\cdot3\sqrt{2}\\ m\amp=1\pm i\cdot3\sqrt{2} \end{align*}

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

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

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Example 7.3.10.

Solve for $y$ in $y^2-4y+13=0\text{,}$ where $y$ might not be a real number.

Explanation

Note that there is a $y$ term, so the square root method is not available. 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 7.3.10, 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.

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Remark 7.3.11.

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 7.3.10, 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:

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

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