P.10. Quadratic Equations
Quadratic equations are an important class of equations which crop up in all areas of math and science. Quadratic equations are equations of the form ax2 + bx + c = 0, where a, b, and c are constants.
There are several methods for solving quadratic equations, three of which we will discuss. The first method is factoring. In Section P.7, we discussed polynomial factorization and in particular, how to factor certain quadratic polynomials p(x) into two linear factors. This method can be applied to solving the quadratic equation p(x) = 0.
Example 1: Solve the quadratic equation x2 - 7x + 10 = 0.
Solution: By trial and error, we find x2 - 7x + 10 = (x - 2)(x - 5). Thus we have (x - 2)(x - 5) = 0. Now if the product of two expressions is zero, then at least one of the expressions must be zero. Thus we have x - 2 = 0, implying x = 2, or x - 5 = 0, implying x = 5. Thus, the two solutions are x = 2 and x = 5. We may check these solutions by plugging them back into the original equation. Plugging in x = 2, we see that the left side of the quadratic equation becomes 22 - (7)(2) + 10 = 4 - 14 + 10 = 0. Substituting x = 5, we check that 52 - (7)(5) + 10 = 25 - 35 + 10 = 0.
A second method of solving quadratic equations is completing the square. We use another example to illustrate this method.
Example 2: Solve the quadratic equation x2 - 6x + 8 = 0.
Solution: Note that by adding 1 to both sides, we obtain a square polynomial on the left, namely x2 - 6x + 9 = 1, which by identity (P.7.2b) becomes (x - 3)2 = 1. Upon computing the square root of both sides, we see that either x - 3 = 1 or x - 3 = -1. The first of these equations yields x = 4 and the second yields x = 2. Thus, the solutions of the quadratic equation are x = 2 and x = 4. It is straightforward to check these solutions.
The third method for solving quadratic equations is by means of a famous formula known as the quadratic formula. The formula for solving the two roots of the quadratic equation ax2 + bx + c = 0 is as follows:
- (P.10.1) x = (-b ± √b2 - 4ac) / 2a.
We illustrate the use of the quadratic formula with a third example.
Example 3: Solve the quadratic equation 2x2 - 5x + 3 = 0.
Solution: The coefficients of the quadratic polynomial are a = 2, b = -5, and c = 3. Applying the quadratic formula, we see that the roots are [-(-5) ± √(-5)2 - (4)(2)(3)] / (2)(2) = (5 ± √25 - 24) / 4 = (5 ± 1) / 4 = 1 or 3/2. Thus, the solutions are x = 1 and x = 3/2. Once again, it is straightforward to check these solutions by plugging them back into the quadratic equation.