Before discussing the method of solution, a clear distinction should be made between the two terms quadratic equation and quadratic function. According to the earlier discussion, the expression P2 + 4P - 5 constitutes a quadratic function, say, f(P). Hence we may write f{P) = P2 + 4P-5 (3.8)

What (3.8) does is to specify a rule of mapping from P to /(F), such as

 p -6 -5 -4 -3 -2 -1 0 1 2 m 7 0 -5 -8 -9 -8 -5 0 7

Although we have listed only nine P values in this table, actually ait the P values in the domain of the function are eligible for listing. It is perhaps for this reason that wc rarely speak of ^solving" the equation f{P) = P2 + 4P - 5, because we normally expect "solution values" to be few in number, but here all P values can get involved. Nevertheless, one may legitimately consider each ordered pair in the table—such as (-6, 7) and (-5, 0)—as a solution of (3.8), since each such ordered pair indeed satisfies that equation. Inasmuch as an infinite number of such ordered pairs can be written, one for each P value, there is an infinite number of solutions to (3.8). When plotted as a curve, these ordered pairs together yield the parabola in Fig. 3.2.

In (3.7), where we set the quadratic function f(P) equal to zero, the situation is fundamentally changed. Since the variable f(P) now disappears (having been assigned a zero value), the result is a quadratic equation in the single variable P) Now that /(P) is

1 The distinction between quadratic function and quadratic equation just discussed can be extended also to cases of polynomials other than quadratic. Thus, a cubic equation results when a cubic function is \$et equal to zero.

FIGURE 3.2

FIGURE 3.2 restricted to a zero value, only a select number of P values can satisfy (3.7) and qualify as its solution values, namely, those P values at which the parabola in Fig. 3.2 intersects the horizontal axis- on which j\P) is zero. Note that this time the solution values are just P values, not ordered pairs. The solution P values are often referred to as the roots of the quadratic equation f(P) = 0, or, alternatively, as the zeros of the quadratic function f{P).

There are two such intersection points in Fig. 3.2, namely, (1,0) and {-5, 0). As required, the second element of each of these ordered pairs {the ordinate of the corresponding point) shows f(P) = 0 in both cases. The first element of each ordered pair (the abscissa of the point), on the other hand, gives the solution value of P. Here we get two solutions»

P* = 1 and PI = -5 but only the first is economically admissible, as negative prices are ruled out. 