ORCCA More on Complex Solutions to Quadratic Equations

Section 11.3 More on Complex Solutions to Quadratic Equations

When we solve a quadratic equation, sometimes there are no real solutions. In this section we will explore when that happens and what it means on a graph. We will also learn how to handle complex solutions algebraically.

Subsection 11.3.1 Applications with Real or Complex Solutions

Let's look at an application where we will determine whether the solutions are real or complex. Iman is a pilot and in a stunt plane performance, she plans to dive the plane toward the ground and then back up. The plane's height can be modeled by a quadratic function. If one possible function is \(h\text{,}\) where \(h(t)=\frac{1}{2}t^2-5t+12\text{,}\) with \(t\) standing for time in seconds after the stunt begins, determine whether the plane would hit the ground during the stunt.

To check whether the plane on that flight path would hit the ground, we will solve the equation \(h(t)=0\text{.}\) We will solve this equation with the quadratic formula. First, we identify that \(a=\substitute{\frac{1}{2}}\text{,}\) \(b=\substitute{-5}\) and \(c=\substitute{12}\text{.}\)

\begin{align*} t\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \amp=\frac{-(\substitute{-5})\pm\sqrt{(\substitute{-5})^2-4(\substitute{\sfrac{1}{2}})(\substitute{12})}}{2(\substitute{\sfrac{1}{2}})}\\ \amp=\frac{5\pm\sqrt{25-24}}{1}\\ \amp=5\pm\sqrt{1}\\ \amp=5\pm1 \end{align*}

So, either:

\begin{align*} t=6\amp\amp\text{ or }\amp\amp t=4 \end{align*}

Figure 11.3.1. Graph of \(y=h(t)\)

This equation has two real solutions and we can see from the graph that the real solutions are the zeros of \(h\text{.}\) The solution \(4\) shows that the plane would hit the ground \(4\) seconds into the stunt, so this is not a good flight path.

To avoid hitting the ground, Iman adjusted the function to \(p\text{,}\) where \(p(t)=\frac{1}{2}t^2-5t+12.5\text{.}\) To see whether the plane on this flight path would hit the ground, we will solve the equation \(p(t)=0\text{.}\) We will again use the quadratic formula to solve this equation. We identify that \(a=\substitute{\frac{1}{2}}\text{,}\) \(b=\substitute{-5}\) and \(c=\substitute{12.5}\text{.}\)

\begin{align*} t\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \amp=\frac{-(\substitute{-5})\pm\sqrt{(\substitute{-5})^2-4(\substitute{\sfrac{1}{2}})(\substitute{12.5})}}{2(\substitute{\sfrac{1}{2}})}\\ \amp=\frac{5\pm\sqrt{25-25}}{1}\\ \amp=5\pm\sqrt{0}\\ \amp=5\pm0\\ \amp=5 \end{align*}

Figure 11.3.2. Graph of \(y=p(t)\)

This equation has one real solution because \(p\) has one zero. This time the plane would hit the ground \(5\) seconds into the stunt. This is also not a good flight path.

Iman again adjusted the flight path to \(q\text{,}\) where \(q(t)=\frac{1}{2}t^2-5t+13\text{.}\) We will solve the equation \(q(t)=0\) using the quadratic formula. Identify that \(a=\substitute{\frac{1}{2}}\text{,}\) \(b=\substitute{-5}\) and \(c=\substitute{13}\text{.}\)

\begin{align*} t\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \amp=\frac{-(\substitute{-5})\pm\sqrt{(\substitute{-5})^2-4(\substitute{\sfrac{1}{2}})(\substitute{13})}}{2(\substitute{\sfrac{1}{2}})}\\ \amp=\frac{5\pm\sqrt{25-26}}{1}\\ \amp=5\pm\sqrt{-1} \end{align*}

Figure 11.3.3. Graph of \(y=q(t)\)

Because the radicand is negative, there are no real solutions and the function has no horizontal intercepts. This means the plane will not touch the ground and Iman can complete her stunt using this path.

In general, the radicand of the quadratic formula, \(b^2-4ac\text{,}\) is called the discriminant. The sign of the discriminant will tell us how many horizontal intercepts the graph of a quadratic function will have

When using the quadratic formula to solve the equation \(h(t)=0\text{,}\) if the discriminant is a positive number, then this tells us that the equation has two real solutions and that the graph of the quadratic function \(h\) has two horizontal intercepts.
When using the quadratic formula to solve the equation \(h(t)=0\text{,}\) if the discriminant is equal to zero, then this tells us that the equation has only one real solution and that the graph of the quadratic function \(h\) has one horizontal intercept.
When using the quadratic formula to solve the equation \(h(t)=0\text{,}\) if the discriminant is a negative number, then this tells us that the equation has no real solutions and that the graph of the quadratic function \(h\) has no horizontal intercepts.

Example 11.3.4.

Use the discriminant to determine how many horizontal intercepts the graphs of each function has.

\(\displaystyle f(x)=3x^2-6x+3\)
\(\displaystyle g(x)=8x^2+3x+2\)
\(\displaystyle h(x)=6x^2+6x-1\)

Explanation

First, we find the values of \(a\text{,}\) \(b\text{,}\) and \(c\text{.}\) In this case, \(a=3\text{,}\) \(b=-6\text{,}\) and \(c=3\text{.}\) Next, we plug these values into the formula for the discriminant, \(b^2-4ac\text{:}\)

\begin{align*} b^2-4ac\amp=(\substitute{-6})^2-4(\substitute{3})(\substitute{3})\\ \amp=36-36\\ \amp=0 \end{align*}

Thus, the discriminant is zero, which means the graph of \(f\) has one horizontal intercept.
First, we find the values of \(a\text{,}\) \(b\text{,}\) and \(c\text{.}\) In this case, \(a=8\text{,}\) \(b=3\text{,}\) and \(c=2\text{.}\) Next, we plug these values into the formula for the discriminant, \(b^2-4ac\text{:}\)

\begin{align*} b^2-4ac\amp=(\substitute{3})^2-4(\substitute{8})(\substitute{2})\\ \amp=9-64\\ \amp=-55 \end{align*}

Thus, the discriminant is negative, which means the graph of \(g\) has no horizontal intercepts.
First, we find the values of \(a\text{,}\) \(b\text{,}\) and \(c\text{.}\) In this case, \(a=6\text{,}\) \(b=6\text{,}\) and \(c=-1\text{.}\) Next, we plug these values into the formula for the discriminant, \(b^2-4ac\text{:}\)

\begin{align*} b^2-4ac\amp=(\substitute{6})^2-4(\substitute{6})(\substitute{-1})\\ \amp=36+24\\ \amp=60 \end{align*}

Thus, the discriminant is positive, which means the graph of \(h\) has two horizontal intercepts.

Example 11.3.5.

Futsal ¹ is a form of what is usually called soccer in the United States. The game is played on a hard court surface and is usually indoors. The ceiling is out of bounds, so if the ball hits the ceiling it goes to the opposing team.

en.wikipedia.org/wiki/Futsal

Borna kicks the ball from the ground with an upward velocity of \(8\) meters per second. The ball's height in meters can be modeled by the quadratic function \(h\text{,}\) where \(h(t)=-4.9t^2+8t\text{,}\) with \(t\) standing for time in seconds after the ball was kicked. If the ceiling height is \(4\) meters, the minimum height allowed by regulation, determine whether the ball will hit the ceiling.

Explanation

To see whether their ball will hit the ceiling, we will solve the equation \(h(t)=4\text{.}\) We could complete the square or use the quadratic formula. Because this equation has decimal coefficients we will use the quadratic formula. We put the equation in standard form and identify that \(a=\substitute{-4.9}\text{,}\) \(b=\substitute{8}\) and \(c=\substitute{-4}\text{.}\)

\begin{align*} -4.9t^2+8t\amp=4\\ -4.9t^2+8t-4\amp=0 \end{align*}

\begin{align*} t\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \amp=\frac{-\substitute{8}\pm\sqrt{\substitute{8}^2-4(\substitute{-4.9})(\substitute{-4})}}{2(\substitute{-4.9})}\\ \amp=\frac{-8\pm\sqrt{64-78.4}}{-9.8}\\ \amp=\frac{-8\pm\sqrt{-14.4}}{-9.8} \end{align*}

The radicand is negative so we can conclude that there are no real solutions to the equation \(h(t)=4\text{.}\) That means the parabola will not cross the line \(y=4\) and the ball will not hit the ceiling.

Example 11.3.6.

Emma kicks the ball from the ground with an upward velocity of \(10\) meters per second. This gives us the quadratic function for the height of the ball \(h(t)=-4.9t^2+10t\text{,}\) with \(t\) standing for time in seconds after the ball was kicked. If the ceiling height is \(4.5\) meters, determine whether the ball will hit the ceiling.

Explanation

To see whether her ball will hit the ceiling, we will solve the equation \(h(t)=4.5\text{.}\) We will use the quadratic formula because this equation has decimal coefficients. We put the equation in standard form and identify that \(a=\substitute{-4.9}\text{,}\) \(b=\substitute{10}\) and \(c=\substitute{-4.5}\text{.}\)

\begin{align*} -4.9t^2+10t\amp=4.5\\ -4.9t^2+10t-4.5\amp=0 \end{align*}

\begin{align*} t\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \amp=\frac{-\substitute{10}\pm\sqrt{\substitute{10}^2-4(\substitute{-4.9})(\substitute{-4.5})}}{2(\substitute{-4.9})}\\ \amp=\frac{-10\pm\sqrt{100-88.2}}{-9.8}\\ \amp=\frac{-10\pm\sqrt{11.8}}{-9.8} \end{align*}

The radicand is positive so there are two real solutions to the equation \(h(t)=4.5\text{.}\) That means the parabola will cross the line \(y=4.5\) and the ball will hit the ceiling.

Subsection 11.3.2 Solving Equations with Complex Solutions

In a physical context we may only want to know whether solutions are real or complex. Or we may want to find the solutions. When the radicand is negative, we need to go into the complex number system. First, we will revisit the definition of complex numbers. Recall that \(i\) is defined as \(\sqrt{-1}\text{.}\)

Definition 11.3.7. Complex Number.

A complex number ² 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.

en.wikipedia.org/wiki/Complex_number

Here are some examples of solving equations that have complex solutions.

Example 11.3.8.

Solve for \(s\) in \(s^2-10s=-34\text{.}\)

Explanation

We will use the method of completing the square. To do so, we need to add \(\left(\frac{b}{2}\right)^2=(-5)^2=25\) to both sides to complete the square.

\begin{align*} s^2-10s\amp=-34\\ s^2-10s\addright{25}\amp=-34\addright{25}\\ (s-5)^2\amp=-9 \end{align*}

\begin{align*} s-5\amp=-\sqrt{-9}\amp\amp\text{or}\amp s-5\amp=\sqrt{-9}\\ s-5\amp=-\sqrt{9}\cdot\sqrt{-1}\amp\amp\text{or}\amp s-5\amp=\sqrt{9}\cdot\sqrt{-1}\\ s-5\amp=-3i\amp\amp\text{or}\amp s-5\amp=3i\\ s\amp=5-3i\amp\amp\text{or}\amp s\amp=5+3i \end{align*}

The solution set is \(\{5-3i, 5+3i\}\text{.}\)

Checkpoint 11.3.9.

The quadratic formula can also be used to solve for complex solutions. Here is an example where it makes more sense to use the quadratic formula.

Example 11.3.10.

Solve for \(x\) in \(5x^2-2x=-3\text{.}\)

Explanation

If we were to complete the square, we would divide both sides by \(5\) and have lots of fractions in our equation. Instead, we will put the equation in standard form and use the quadratic formula.

\begin{align*} 5x^2-2x\amp=-3\\ 5x^2-2x+3\amp=0 \end{align*}

We identify that \(a=\substitute{5}\text{,}\) \(b=\substitute{-2}\) and \(c=\substitute{3}\) and substitute them into the Quadratic Formula:

\begin{align*} x\amp=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \amp=\frac{-(\substitute{-2})\pm\sqrt{(\substitute{-2})^2-4(\substitute{5})(\substitute{3})}}{2(\substitute{5})}\\ \amp=\frac{2\pm\sqrt{4-60}}{10}\\ \amp=\frac{2\pm\sqrt{-56}}{10}\\ \amp=\frac{2\pm\sqrt{-1\cdot4\cdot14}}{10}\\ \amp=\frac{2\pm\sqrt{-1}\cdot\sqrt{4}\cdot\sqrt{14}}{10}\\ \amp=\frac{2\pm i\cdot2\cdot\sqrt{14}}{10} \end{align*}

Now we need to put the solutions in standard form, which is \(a+bi\text{.}\)

\begin{align*} x\amp=\frac{2}{10}\pm \frac{2i\sqrt{14}}{10}\\ x\amp=\frac{1}{5}\pm \frac{\sqrt{14}}{5}i \end{align*}

The solution set is \(\left\{\frac{1}{5}- \frac{\sqrt{14}}{5}i,\frac{1}{5}+ \frac{\sqrt{14}}{5}i\right\}\text{.}\)

Exercises 11.3.3 Exercises

Review and Warmup

1.

Simplify the radical and write it as a complex number using \(i\text{.}\)

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

2.

Simplify the radical and write it as a complex number using \(i\text{.}\)

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

3.

Simplify the radical and write it as a complex number using \(i\text{.}\)

\(\displaystyle{ \sqrt{-90} =}\)

4.

Simplify the radical and write it as a complex number using \(i\text{.}\)

\(\displaystyle{ \sqrt{-24} =}\)

5.

Simplify the radical and write it as a complex number using \(i\text{.}\)

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

6.

Simplify the radical and write it as a complex number using \(i\text{.}\)

\(\displaystyle{ \sqrt{-180} =}\)

Real Versus Complex Solutions

7.

Determine the nature of the solutions to this quadratic equation.

\(\displaystyle{{-3r^{2}+3r-8} = {0}}\)

two real solutions
two non-real solutions
one doubled real solution
none of these

8.

Determine the nature of the solutions to this quadratic equation.

\(\displaystyle{{-10r^{2}+20r+7} = {0}}\)

two real solutions
two non-real solutions
one doubled real solution
none of these

9.

Determine the nature of the solutions to this quadratic equation.

\(\displaystyle{{4x^{2}-3x+1} = {0}}\)

two real solutions
two non-real solutions
one doubled real solution
none of these

10.

Determine the nature of the solutions to this quadratic equation.

\(\displaystyle{{-4x^{2}+14x-5} = {0}}\)

two real solutions
two non-real solutions
one doubled real solution
none of these

11.

Determine the nature of the solutions to this quadratic equation.

\(\displaystyle{{10y^{2}-3y-10} = {0}}\)

two real solutions
two non-real solutions
one doubled real solution
none of these

12.

Determine the nature of the solutions to this quadratic equation.

\(\displaystyle{{3z^{2}+2z+5} = {0}}\)

two real solutions
two non-real solutions
one doubled real solution
none of these

13.

Determine the nature of the solutions to this quadratic equation.

\(\displaystyle{{-5z^{2}-5z-1} = {0}}\)

two real solutions
two non-real solutions
one doubled real solution
none of these

14.

Determine the nature of the solutions to this quadratic equation.

\(\displaystyle{{9t^{2}-4t+7} = {0}}\)

two real solutions
two non-real solutions
one doubled real solution
none of these