ORCCA Complex Solutions to Quadratic Equations

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.

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

(1 - t) (3 - t) + 10 = 0

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Let’s practice solving this equation.

\begin{aligned} (1 - t) (3 - t) + 10 & = 0 \\ 3 - t - 3 t + t^{2} + 10 & = 0 \\ t^{2} - 4 t + 13 & = 0 \end{aligned}

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We can try the quadratic formula.

\begin{aligned} t & = \frac{4 \pm \sqrt{(- 4)^{2} - 4 (1) (13)}}{2 (1)} \\ = \frac{4 \pm \sqrt{16 - 52}}{2} \\ = \frac{4 \pm \sqrt{- 36}}{2} \\ = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{36}}{2} \\ = \frac{4 \pm i \cdot 6}{2} \\ = 2 \pm 3 i \end{aligned}

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These two solutions,

2 - 3 i

and

2 + 3 i

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.

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

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Example 7.3.9.

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Solve for

m

(m - 1)^{2} + 18 = 0,

where

m

might not be a real number.

Explanation.

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

\begin{aligned} (m - 1)^{2} + 18 & = 0 \\ (m - 1)^{2} & = - 18 \\ m - 1 & = \pm \sqrt{- 18} \\ m - 1 & = \pm \sqrt{- 1 \cdot 9 \cdot 2} \\ m - 1 & = \pm \sqrt{- 1} \cdot \sqrt{9} \cdot \sqrt{2} \\ m - 1 & = \pm i \cdot 3 \sqrt{2} \\ m & = 1 \pm i \cdot 3 \sqrt{2} \end{aligned}

\begin{aligned} m & = 1 - 3 i \sqrt{2} & or & m & = 1 + 3 i \sqrt{2} \end{aligned}

The solution set is

{1 - 3 i \sqrt{2}, 1 + 3 i \sqrt{2}} .

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Example 7.3.10.

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Solve for

y

y^{2} - 4 y + 13 = 0,

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,

b = - 4

and

c = 13

and substitute them into the quadratic formula.

\begin{aligned} y & = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ = \frac{- (- 4) \pm \sqrt{(- 4)^{2} - 4 (1) (13)}}{2 (1)} \\ = \frac{4 \pm \sqrt{16 - 52}}{2} \\ = \frac{4 \pm \sqrt{- 36}}{2} \\ = \frac{4 \pm \sqrt{- 1} \cdot \sqrt{36}}{2} \\ = \frac{4 \pm 6 i}{2} \\ = 2 \pm 3 i \end{aligned}

The solution set is

{2 - 3 i, 2 + 3 i} .

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Note that in Example 10, the expressions

2 + 3 i

and

2 - 3 i

are fully simplified. In the same way that the terms

2

and

3 x

cannot be combined, the terms

2

and

3 i

can not be combined.

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Remark 7.3.11.

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Each complex solution can be checked, just as every real solution can be checked. For example, to check the solution of

2 + 3 i

from Example 10, we would replace

y

with

2 + 3 i

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 .

This check is shown here:

\begin{aligned} y^{2} - 4 y + 13 & = 0 \\ (2 + 3 i)^{2} - 4 (2 + 3 i) + 13 & \overset{?}{=} 0 \\ (2^{2} + 2 (3 i) + 2 (3 i) + (3 i)^{2}) - 4 \cdot 2 - 4 \cdot (3 i) + 13 & \overset{?}{=} 0 \\ 4 + 6 i + 6 i + 9 i^{2} - 8 - 12 i + 13 & \overset{?}{=} 0 \\ 4 + 9 (- 1) - 8 + 13 & \overset{?}{=} 0 \\ 4 - 9 - 8 + 13 & \overset{?}{=} 0 \\ 0 & \overset{✓}{=} 0 \end{aligned}

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Reading Questions 7.3.4 Reading Questions

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1.

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What is

{(i^{2})}^{2} ?

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2.

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A number like

4 i

is called a number. A number like

3 + 4 i

is called a number.

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Simplify the radical and write it as a complex number using

i .

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\sqrt{- 70} =

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2.

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Simplify the radical and write it as a complex number using

i .

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\sqrt{- 105} =

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3.

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Simplify the radical and write it as a complex number using

i .

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\sqrt{- 27} =

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4.

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Simplify the radical and write it as a complex number using

i .

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\sqrt{- 54} =

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5.

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Simplify the radical and write it as a complex number using

i .

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\sqrt{- 135} =

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6.

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Simplify the radical and write it as a complex number using

i .

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\sqrt{- 270} =

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