Complex Roots of a Quadratic Equation Graphically.

Fig. 1. Graphic solution of x2 + ax + b = 0 when the roots are complex. C: y = 2 + a + 6. To find the complex roots a ? i? of (1), proceed as follows: Draw the axis of sym metry of the parabola and mark points V the vertex of the parabola and the inter section of the axis of the parabola with the x-axis. Determine Q so that V is the mid point of segment PQ. Draw the line paral lel to the x-axis passing through Q and intersecting the parabola in points R and S. We shall show a and ? to be the x-coor dinates of and | QS\ respectively. First by the quadratic formula the roots of (1) are (? a ? Va2 ? 46)/2 and, since a2 ? 46 < 0, the roots may be rewritten as ? a/2 ? t'(\/46 ? a2)/2. Hence, a = a/2 and ? = (V46 a2)/2. Now the parabola C: y = 2 + a + 6 of figure 1 has standard form y (6 a2/4) = ( + a/2)2. Thus and V will have coordinates (? a/2, 0), (? a/2, 6 ? a2/4) respectively, and Q will have coordinates (? a/2, 26 ? a2/2). Thus a = ? a/2 is verified as the x-coor dinate of (or V or Q). The fact that ? = I QS\ remains to be verified. The x-coordinate of points R and S will be found by solving the equations of C and RS simultaneously. The line RS will have equation y = 26 ? a2/2. S and R will have, as x-coordinates, the solutions of the equation x2 + ax + 6 = 26 ? a2/2, i.e., a/2 ? (V46 a2)/2 and | QS\ =V462 a2/2. Then ? = ?(?Si The above method gives the complex roots almost as easily as the conventional graphic method yields the real roots.