That’s the left side of the original quadratic. Now we know that it can be written as
(x+ 3)(x− 5) = 0
The roots are easy to find. If x=−3 or x= 5, the left side of the equation becomes equal to 0. The solution
setX is therefore
X= {−3,5}
Let’s check to be sure these solutions work in the original equation. For x=−3, we have
x^2 − 2 x− 15 = 0
(−3)^2 − 2 × (−3)− 15 = 0
9 − (−6)− 15 = 0
9 + 6 − 15 = 0
15 − 15 = 0
0 = 0
For x= 5, we have
x^2 − 2 x− 15 = 0
52 − 2 × 5 − 15 = 0
25 − 10 − 15 = 0
15 − 15 = 0
0 = 0
Completing the Square
There are other ways to look for the roots of a quadratic. One of these methods is called com-
pleting the square.
Perfect squares
Suppose that we come across a quadratic equation whose left side breaks down into two
identical factors, equivalent to a single factor multiplied by itself. A polynomial of this type is
called a perfect square. Here are some examples:
- The polynomial x^2 + 2 x+ 1 factors into (x+ 1)^2
- The polynomial x^2 − 2 x+ 1 factors into (x− 1)^2
- The polynomial x^2 + 4 x+ 4 factors into (x+ 2)^2
- The polynomial 9x^2 + 12 x+ 4 factors into (3x+ 2)^2
Quadratics built from perfect squares (by setting the right side of the equation equal to 0) are
easy to solve. The two roots are identical, so in effect there’s only one root, “done twice over.”
Completing the Square 371
