Theorem: (x + r)(x + s) = x2 + (r + s)x + rs.
Proof:
| (x + r)(x + s) = x2 + xs + rx + rs | (Distributive Law) |
| = x2 + sx + rx + rs | (Commutative Law) |
| = x2 + (r + s)x + rs | (Distributive Law) |
We require both the distributive law and the commutative law of multiplication for this proof. There aren’t many interesting algebraic structures without the distributive law; it is too fundamental. But we will have examples shortly of structures that don’t satisfy the commutative law, and it is important to note that our basic rules of factoring won’t work for such structures.
To apply this idea, suppose you want to solve the equation x2 + x + 2 = 0 (mod 11). We want to factor this by finding values for r and s so r + s = 1 and rs = 2. If this seems impossible, you are thinking about real numbers. Remember that we are working in Z11. In this case, 5 + 7 = 1 (mod 11) and 5 ´ 7 = 2 (mod 11), so we can indeed factor x2 + x + 2 = (x + 5)(x + 7). Now we can set (x + 5)(x + 7) = 0. In a field like Z11, the only way a product can equal 0 is if one of the factors is 0. Since we proved earlier that no zero divisor has an inverse, this is a consequence of law 8 on the existence of multiplicative inverses. So we can reduce our problem to x + 5 = 0 or x + 7 = 0, and the roots are -5 and -7, or 6 and 4 (mod 11).
The other approach to solving quadratic equations is to use the quadratic formula. This formula also works for general fields. The key idea is completing the square, which follows from the rules for factoring we’ve already verified.
Theorem: The solutions to ax2 + bx + c = 0 are -b´(2a)-1 ± Ö((b´(2a)-1)2 - c´a-1 ).
Proof: We start by multiplying the original equation by a-1 to get x2 + (b´a-1)x + c´a-1 = 0. Then we complete the square to get (x + b´(2a)-1)2 + c´a-1 - (b´(2a)-1)2 = 0. Next we move the constant terms to the other side of the equation yielding (x + b´(2a)-1)2 = (b´(2a)-1)2 - c´a-1. Then we take square roots of both sides of the equation to get x + b´(2a)-1 = ±Ö((b´(2a)-1)2 - c´a-1). Finally we solve for x to get our formula x = -b´(2a)-1 ± Ö((b´(2a)-1)2 - c´a-1 ).
This is the same as the usual quadratic formula. It may look different from the way you learned it because we refer to multiplicative inverses rather than simply dividing and because I haven’t factored out a 2a term for the denominator (there being no denominator since I don’t have fractions). You should be aware that the quadratic formula can be expressed in a variety of different forms, and schools in other countries often use different forms than we use. Especially if you teach in a town like Manhattan, where you have children of foreign faculty and students and children of military personnel who have been stationed overseas, you should be prepared for students who have learned the quadratic formula in a different form from that conventionally taught in this country.
To apply the quadratic formula to our earlier problem, x2 + x + 2 = 0 (mod 11), we note that a = 1, b = 1, and c = 2. So our formula gives x = -2-1 ± Ö((2-1)2 - 2). Now 2-1 = 6 (mod 11), so x = -6 ± Ö(62 - 2) = 6 ± Ö(34) = -6 ± Ö(1) (mod 11), so our solutions are -5 and -7, or 6 and 4 (mod 11), just as we found before.
I have a confession to make. I cheated in the development of the quadratic formula. I used square roots and we haven’t discussed anything about square roots in general fields. How do we know that square roots come in positive and negative pairs for general fields? Of course this all works out; otherwise I wouldn’t have bothered teaching the quadratic formula. But the facts we need about square roots need justification. That is where we will begin the next handout.