5.4 The Quadratic Formula - College of the Redwoods

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492 Chapter 5 Quadratic Functions y ( √ ) ( √ ) 4− 32 2 , 0 4+ 32 2 , 0 x (2, −8) Figure 2. If the discriminant is positive, there are two real x-intercepts. Thus, if the discriminant is positive, the parabola will have two real x-intercepts. Next, let’s look at an example where the discriminant equals zero. ◮ Example 16. Consider again the quadratic equation ax 2 + bx + c = 0 and the solutions (zeros) provided by the quadratic formula x = −b ± √ b 2 − 4ac . 2a The expression under the radical, b 2 − 4ac, is called the discriminant, which we denote by the letter D. That is, the formula for the discriminant is given by D = b 2 − 4ac. The discriminant is used to determine the nature and number of solutions to the quadratic equation ax 2 +bx+c = 0. This is done without actually calculating the solutions. Consider the quadratic equation x 2 − 4x + 4 = 0. Calculate the discriminant and use it to determine the nature and number of the solutions. Compare x 2 − 4x + 4 = 0 with ax 2 + bx + c = 0 and note that a = 1, b = −4, and c = 4. The discriminant is given by the calculation Note that the discriminant equals zero. D = b 2 − 4ac = (−4) 2 − 4(1)(4) = 0. Consider the quadratic function f(x) = x 2 − 4x + 4, which can be written in vertex form Version: Fall 2007
Section 5.4 The Quadratic Formula 493 f(x) = (x − 2) 2 . (17) This is a parabola that opens upward and is shifted 2 units to the right. Note that there is no vertical shift, so the vertex of the parabola will rest on the x-axis, as shown in Figure 3. In this case, we found it necessary to plot two points to the right of the axis of symmetry, then mirror them across the axis of symmetry, in order to get an accurate plot of the parabola. y 10 (2, 0) x 10 x f(x) = (x − 2) 2 3 1 4 4 x = 2 Figure 3. At the right is a table of points satisfying f(x) = (x−2) 2 . These points and their mirror images are seen as solid dots superimposed on the graph of f(x) = (x − 2) 2 at the left. Take a closer look at equation (17). If we set f(x) = 0 in this equation, then we get 0 = (x − 2) 2 . This could be written 0 = (x − 2)(x − 2) and we could say that the solutions are 2 and 2 again. However, mathematicians prefer to say that “2 is a solution of multiplicity 2” or “2 is a double solution.” 11 Note how the parabola is tangent to the x-axis at the location of the “double solution.” That is, the parabola comes down from positive infinity, touches (but does not cross) the x-axis at x = 2, then rises again to positive infinity. Of course, the situation would be reversed in the parabola opened downward, as in g(x) = −(x − 2) 2 , but the graph would still “kiss” the x-axis at the location of the “double solution.” Still, the key thing to note here is the fact that the discriminant D = 0 and the parabola has only one x-intercept. That is, the equation x 2 − 4x + 4 = 0 has a single real solution. Next, let’s look what happens when the discriminant is negative. ◮ Example 18. Consider the quadratic equation x 2 − 4x + 8 = 0. 11 Actually, mathematicians call these “double roots,” but we prefer to postpone that language until the chapter on polynomial functions. Version: Fall 2007
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5.4 The Quadratic Formula - College of the Redwoods

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