###
Subsection 12.3.1 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 12.3.2 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 12.3.3 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 12.3.4 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, \(b^2-4ac\) is called the discriminant. The sign of the discriminant will tell us how many horizontal intercepts a quadratic function will have

When a quadratic function \(h\) has two horizontal intercepts, the equation \(h(t)=0\) has two real solutions. The discriminant will be a positive number so that the \(\pm\) from the quadratic formula will provide two solutions.

When a quadratic function \(p\) has one horizontal intercept, the equation \(p(t)=0\) has one real solution. The discriminant will be zero so that the \(\pm\) from the quadratic formula will provide only one solution.

When a quadratic function \(q\) has no horizontal intercepts, the equation \(q(t)=0\) has no real solutions, but it has two complex solutions. The discriminant will be a negative number so that the \(\sqrt{\phantom{\Delta}}\) from the quadratic formula will provide imaginary numbers, and then the \(\pm\) will provide two complex solutions.

######
Example 12.3.5

Futsal^{β1β} 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.

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 12.3.6

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 12.3.2 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 12.3.7 Complex Number

A complex number^{β2β} is a number that can be expressed in the form \(a + bi\), 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.

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

######
Example 12.3.8

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 \(\{5-3i, 5+3i\}.\)

######
Checkpoint 12.3.9

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 12.3.10

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 \(\left\{\frac{1}{5}- \frac{\sqrt{14}}{5}i,\frac{1}{5}+ \frac{\sqrt{14}}{5}i\right\}.\)