5.4 The Quadratic Formula - College of the Redwoods

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488 Chapter 5 Quadratic Functions Solution Strategy—Linear Versus Nonlinear. When solving equations, you must first ask if the equation is linear or nonlinear. Again, let’s assume the unknown we wish to solve for is x. • If the equation is linear, move all terms containing x to one side of the equation, all the remaining terms to the other side of the equation. • If the equation is nonlinear, move all terms to one side of the equation, making the other side of the equation zero. Thus, because ax + b = cx + d is linear in x, the first step in solving the equation would be to move all terms containing x to one side of the equation, all other terms to the other side of the equation, as in ax − cx = d − b. On the other hand, the equation ax 2 + bx = cx + d is nonlinear in x, so the first step would be to move all terms to one side of the equation, making the other side of the equation equal to zero, as in ax 2 + bx − cx − d = 0. In Example 13, the equation x 2 −2x = 2 is nonlinear in x, so we moved everything to the left-hand side of the equation, making the right-hand side of the equation equal to zero, as in x 2 − 2x − 2 = 0. However, it doesn’t matter which side you make equal to zero. Suppose instead that you move every term to the right-hand side of the equation, as in 0 = −x 2 + 2x + 2. Comparing 0 = −x 2 + 2x + 2 with general quadratic equation 0 = ax 2 + bx + c, note that a = −1, b = 2, and c = 2. Write down the quadratic formula. x = −b ± √ b 2 − 4ac 2a Next, substitute a = −1, b = 2, and c = 2. Again, note the careful use of parentheses. This leads to two solutions, x = −(2) ± √ (2) 2 − 4(−1)(2) 2(−1) x = −2 ± √ 4 + 8 −2 = −2 ± √ 12 . −2 In Example 13, we found the following solutions and their approximations. x = 2 − √ 12 2 ≈ −0.7320508076 and x = 2 + √ 12 2 ≈ 2.732050808. Version: Fall 2007
Section 5.4 The Quadratic Formula 489 It is a fair question to ask if our solutions x = (−2 ± √ 12)/(−2) are the same. One way to find out is to find decimal approximations of each on our calculator. x = −2 − √ 12 −2 ≈ 2.732050808 and x = −2 + √ 12 −2 ≈ −0.7320508076. The fact that we get the same decimal approximations should spark confidence that we have the same solutions. However, we can also manipulate the exact forms of our solutions to show that they match the previous forms found in Example 13. Take the two solutions and multiply both numerator and denominator by minus one. −2 − √ 12 −2 = 2 + √ 12 2 and −2 + √ 12 −2 = 2 − √ 12 2 This shows that our solutions are identical to those found in Example 13. We can do the same negation of numerator and denominator in compact form. −2 ± √ 12 −2 = 2 ∓ √ 12 2 Note that this leads to the same two answers, (2 − √ 12)/2 and (2 + √ 12)/2. Of the two methods (move all the terms to the left or all the terms to the right), we prefer the approach of Example 13. By moving the terms to the left-hand side of the equation, as in x 2 − 2x − 2 = 0, the coefficient of x 2 is positive (a = 1) and we avoid the minus sign in the denominator produced by the quadratic formula. Intercepts In Example 13, we used the quadratic formula to find the solutions of x 2 − 2x − 2 = 0. These solutions, and their approximations, are shown in equation (14). It is important to make the connection that the solutions in equation (14) are the zeros of the quadratic function g(x) = x 2 − 2x − 2. The zeros also provide the x-coordinates of the x-intercepts of the graph of g (a parabola). To emphasize this point, let’s draw the graph of the parabola having the equation g(x) = x 2 − 2x − 2. First, complete the square to place the quadratic function in vertex form. Take half the middle coefficient and square, as in [(1/2)(−2)] 2 = 1; then add and subtract this term so the equation remains balanced. g(x) = x 2 − 2x − 2 g(x) = x 2 − 2x + 1 − 1 − 2 Factor the perfect square trinomial, then combine the constants at the end. g(x) = (x − 1) 2 − 3 This is a parabola that opens upward. It is shifted to the right 1 unit and down 3 units. This makes it easy to identify the vertex and draw the axis of symmetry, as shown in Figure 1(a). Version: Fall 2007
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5.4 The Quadratic Formula - College of the Redwoods

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