ORCCAOpen Resources for Community College Algebra

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*}

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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*}

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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*}

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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.

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 \(\sqrt{-1}\text{.}\)

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

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.