When solving a quadratic equation, follow these steps (in this order) to decide on a method:
Example 1: Solve x^{2} + 4 = 4x
First, put the equation in standard form so that we can try to solve it by factoring:
x^{2} - 4x + 4 = 0
(x - 2)(x - 2) = 0
x - 2 = 0 | x - 2 = 0
x = 2 | x = 2
So the solution to this equation, found by factoring,
is x = 2.
Example 2: Solve (2x - 2)^{2} = -4
The side of the equation containing the variable (the left side) is a perfect square, so we'll take the square root of both sides to solve the equation.
(2x - 2)^{2} = -4
2x - 2 = ± 2i
2x = 2 ± 2i
x = 1 ± i
Notice that the ± sign
was inserted in the equation at the point that the square root was taken.
Example 3: Solve x^{2} + 6x - 11 = 0
This equation is not factorable, and the side containing the variable is not a perfect square. But since the coefficient of the x^{2} is 1 and the coefficient of the x is even, completing the square will be an appropriate method. To find the number which needs to be added to both sides of the equation to complete the square, take the coefficient of the x term, divide it by 2, then square that number. In this problem, 6 ¸ 2 is 3, and 3^{2} is 9, so we'll add 9 to both sides of the equation once we have isolated the variable terms.
x^{2} + 6x - 11 = 0
x^{2} + 6x = 11
x^{2} + 6x +9 = 11 + 9
(x + 3)^{2} = 20
Example 4: Solve 2x^{2} - x + 5 = 0
This equation is not factorable, the left side is not a perfect square, and the coefficients of the x^{2} and x terms will not make completing the square convenient. That leaves the quadratic formula as the best method for solving this equation. We'll use a=2, b=-1, and c=5.