11.3.6 The General Quadratic Equation When we put all of the complexities of Eq. (5) and Eq. (9) together we can write the most general form for a quadratic equation, namely: (y + y0) = a(x + x0)2 (10) We want to find values for x when y = 0. Thus, set y = 0 into Eq. (10) and solve for x: (0 + y0) = a(x + x0)2 y0 = a(x + x0)2 0 = a(x2 + 2x0* x + x02) – y0 0 = ax2 + 2ax0* x + (ax02 – y0) (11) Recognizing that the terms 2ax0 and (ax02 – y0) are composed of coefficients and constants, and not variables, we define a simpler set of coefficients, a and b, and the constant, c: b = 2ax0 and c = ax02 – y0.
Substituting for the constants b and c back into Eq. (11), we have: 0 = ax2 + b x + c, (12) This is the quadratic equation whose general solutions are detailed in the next section.
11.4 Solutions to The Quadratic Equation We can encounter equations with variables, coefficients and constants that are related through any arithmetic operation discussed so far. To find a solution to Eq. (12) means to place onto one side of the equals sign the variable of interest, raised to the first power and with a coefficient of 1, and place on the other side of the equals sign all the other variables, coefficients and constants. To obtain the first power of x, we want the lefthand side to be a perfect square. As we have seen in Section 11.3.6, the quadratic equation is defined by Eq. (12): ax2 + bx + c = 0. (12) It contains three terms: 1) the variable x, raised to the second power and multiplied by the quadratic coefficient, a; 2) the variable x, raised to the first power and multiplied by the linear coefficient, b; and 3) the constant, c.
If the quadratic coefficient a were zero, Eq. (12) would reduce to: 0 + bx + c = 0, which is a linear equation with general solution (b≠0): x = c/b. Linear relationships were discussed in Chapter 10 [See, for example, Section 10.3] and are not the focus of discussion here. Therefore, in what follows, we require that a ≠ 0 in Eq. (12) and Eq. (13a), and we seek a perfect square for the lefthand side of Eq. (13a).
x2 + (b/a)x = –c/a. (13a)
11.4.1 A Graphical Method for Completing the Square We present two methods of completing the square: one literally and graphically completes the square, and the other analytically completes the square. First consider the graphical method shown in Fig. 10. Fig. 10 A graphical method or completing the square.
The quadratic term on the left of Eq. (13a) is x2. This term is graphed in Fig. 10 as an x by x square. The linear term (b/a)x is graphically represented by a rectangle, one of whose sides is x and whose other side is (b/a). Cut this rectangle in half forming areas A and B, each with sides x and (b/2a). Take the B rectangle and place it under the square whose sides are x, as shown in Fig. 10. When we add the square on the lower right, with sides of (b/2a), and area of (b/2a)2, we have completed the square, whose larger sides are: [x + (b/2a)].
Referring to Eq. (13a), we see that if we add (b/2a)2 to its right hand side, we will construct a perfect square on the left given by [x + (b/2a)]2. Specifically: [x + (b/2a)]2 = –c/a + (b/2a)2 Rearranging the right hand side we find: [x + (b/2a)]2 = (b/2a)2 – c/a. 11.4.2 An analytic method for completing the square There exists a general analytic solution for Eq. (12), and we shall now find it. Eq. (12) is: ax2 + bx + c = 0. (12) Divide both sides of this equation by a to find: x2 + (b/a)x + c/a = 0. Subtract c/a from both sides of the equation: x2 + (b/a)x + c/a  c/a = 0 – c/a, x2 + (b/a)x + 0 = – c/a, or x2 + (b/a)x = –c/a. (13a)
What arithmetic operations must be done to put Eq. (13a) into the form of Eq. (1), a perfect square? (z + e)2 = z2 + 2ez + e2. (1)
Comparing liketerms between Eq. (13a) and Eq. (1), we find the second, linear term in Eq. (13a) requires a 2 as part of the linear coefficient, 2ez. So we multiply the second term in Eq. (13a) by 2/2 (=1):
x2 + 2(b/2a)x = –c/a. (13b)
Continuing the correspondence, we identify the variable, z, in Eq. (1) with the variable, x, in Eq. (13b). The term, e, in Eq. (1) is identified with the term, (b/2a), in Eq. (13b). Then (b/2a)2 must correspond to e2.
This means if we add the term (b/2a)2 to both sides of Eq. (13b), we will transform the left hand side of the equation into a perfect square.
x2 + 2(b/2a)x + (b/2a)2 = + (b/2a)2 – c/a. (13c)
The left hand side of Eq. (13c) is of the form of Eq. (1), and so we can write:
[x + (b/2a)]2 = (b/2a)2 – c/a. (13d)
To find x, we take the square root of both sides of Eq. (13d):
x + (b/2a) = ±[(b/2a)2 – c/a]1/2. (13e)
The ± sign shows that there are two possible solutions when taking the square root. [See Section 8.5.] We isolate the variable, x, by Subtracting (b/2a) from both sides of Eq. (13e). We find the solutions for x are: x + (b/2a)  (b/2a) = –(b/2a) ±[(b/2a)2  c/a]1/2, x + 0 = –(b/2a) ±[(b/2a)2  c/a]1/2, or x = – (b/2a) ±[(b/2a)2 – c/a]1/2. (13f)
We now collect like terms on the right hand side of Eq. (13f) to reduce the two solutions to their simplest terms. The terms under the square root define the discriminate (See Section 9.4): (b/2a)2 – c/a. Carrying out the squaring operation indicated in the first term, we have: b2/4a2 – c/a. To find a common denominator we multiply the c/a term by 4a2/4a2 (=1): b2/4a2 – (c/a) * (4a2/4a2) b2/4a2 – (4ac/4a2) Factor the common denominator: 1/4a2 to find that the terms under the square root are equal to:
(1/4a2)(b2 – 4ac). Placing this expression back under the square root in Eq. (13f), we have:
x = – (b/2a) ± [(1/4a2)(b2 – 4ac)]1/2.
x = – (b/2a) ± (1/4a2)1/2 *[(b2 – 4ac)]1/2.
