Review of Complex Numbers and Functions

The Quadratic Formula

Equations of the form

$\begin{displaymath} ax^2 + bx + c = 0 \end{displaymath}$ (1)

are called quadratic equations. The quadratic formula

$\begin{displaymath} x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} \end{displaymath}$ (2)

gives the solutions.
The $\pm$ in Eq. 2 is an important detail. In general, there are two solutions to a quadratic equation. The two solutions are also called the roots of the equation.
Example
A ball is tossed directly upward from a height of 2.0 m above the ground with an initial velocity of 5.0 m/s. It is subject only to the force of gravity while in flight, so it has an acceleration of -9.8 m/s. When does the ball reach a height of 3.0 m? The position of the ball as a function of time is given by the equation

$\begin{displaymath} y = y_o + v_o t + \frac{1}{2}a t^2. \end{displaymath}$

Solution
The equation with the given information is

$\begin{displaymath} 3.0~\mathrm{m} = 2.0~\mathrm{m} + (5.0~\mathrm{m/s}) t + (-4.9~\mathrm{m/s}^2) t^2 \end{displaymath}$

which in the form of Eq. 1 becomes

$\begin{displaymath} (-4.9~\mathrm{m/s}^2) t^2 + (5.0~\mathrm{m/s}) t + (-1.0~\mathrm{m}) = 0. \end{displaymath}$ (3)

We can identify , , and and apply Eq. 2,

$\begin{eqnarray*} t &=& -\frac{5.0~\mathrm{m/s}}{2(-4.9~\mathrm{m/s}^2)} \pm \f... ...{m/s}^2)}\\ &=& 0.51~s \mp 0.24~s = \framebox{0.27~s, 0.75~s}. \end{eqnarray*}$

To understand the two solutions, a graph of the left side of Eq. 3 is helpful
$\includegraphics{proj.eps}$
- The extreme value of the quadratic occurs at $-\frac{b}{2a}$ . In the example, the extreme value is at 0.51 s.
- The roots of the quadratic lie $\frac{\sqrt{b^2 - 4ac}}{2a}$ to either side of $-\frac{b}{2a}$ . In the example, the two roots of the equation lie 0.24 s to either side of 0.51 s.
- A graph like this is often helpful in choosing appropriate solutions. In the example, both roots of the equation are appropriate. The ball reaches a height of 3.0 m twice, once going up, and once coming down.
Special Cases
- Single solution
  If , then $\sqrt{b^2-4ac} = 0$ , and there is only one solution, $x=-\frac{b}{2a}$ . In the above example, this corresponds to 0.51 s, the time at which the ball reaches its maximum height.
- Complex solutions
  If , then $\sqrt{b^2-4ac}$ is imaginary (involves $i = \sqrt{-1}$ ). The above example does not have a physically reasonable solution corresponding to this situation. (This situation corresponds to heights never reached by the ball.)