140. A quadratic equation having real roots may be solved graphically by the principles of graphs previously explained. 1. Let it be required to solve x2 + x − 6 = 0. If we let x2 + x - 6: found. = y, the graph of this equation may be Substituting values for x, we find the corresponding values of y. Locate the various points as shown in Fig. I, and connect these points by a smooth curve, which represents the graph of x2 + x == y. ་ The roots of the equation, x2 + x 6 = 0, are those values of x 0, but since y = x2 + x that will make x2 + x 6 = 6, any value of x that makes y equal to zero, must be a root of the equation. The ordinate y becomes zero for any point on the X axis, hence the roots are located by the intersections of the graph with the X axis. In Fig. I, the graph cuts the X axis, when x = 2 and again when x = -3, therefore the roots are 2 and -3. Since the value of an algebraic expression, like x2 + x - 6, depends upon the value of the letter involved, it is called a function of that letter. Thus, x2 x 6 is a function of x, expressed by the notation, f(x). Hence, since x2 + x 6 = f(x), Any value of x that makes f(x) equal to zero will be a root of the equation x2 + x − 6 = 0. In the graphical solution of this equation, the graph shows the corresponding values of x and f(x), that is, the abscissa of any point is the value of x, and the corresponding ordinate is the value of f(x). Plotting the graph of x2 4x + 4 =y, Fig. II, we note that the curve touches the X axis at +2, but does not cross the axis. This indicates that the two roots of the equation are each equal to +2. graph approaches the X axis but does not cut it. This shows that the equation x2 + 4x + 6 O has no real roots, both roots being imaginary, as may be shown by an algebraic solution. = X FIG. IV. The graph of any equation of the form ax2 + bx + c = O as shown, is one of the curves produced by conic sections and is called a parabola. Note. When the roots of an equation are fractional, the graph will not intersect the X axis at a point of division and the values of the roots must be estimated approximately. In such cases, the graph should be drawn as accurately as possible. A modification of the foregoing method is sometimes used. 4. Solve x2 + x 6 = 0. Let x2 (1) Using these two equations as a simultaneous system, the graph of equation (1) is a parabola, and that of equation (2) a straight line, Fig. IV. |
