17. Solving Quadratic Equations by Factoring
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Learning Objective:
Apply factoring, if possible, to solve a quadratic equation in one variable by finding the root(s) of the equation, if any, i.e., the value(s) of the variable that make the value of the quadratic function equal to zero.
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Quadratic equations in one variable
Define a quadratic equation in one variable as ax² + bx + c = 0, where a, b, and c are real numbers and a ≠ 0. Solving this equation means to find all the values of x that satisfy this equation, i.e., the left-hand and right-hand sides of the equation are equal. This is also known as finding the roots of the equation.
Recall that the zero-factor theorem says that if a product of two numbers (factors) is 0, then one or both factors must be 0. Also recall that sometimes it is possible to factor a quadratic. We can put these two ideas together to easily solve some quadratic equations.
Two solutions
For example, 2x – 6 = 0 ⇒ 2x – 6 + 6 = 0 + 6 ⇒ 2x = 6 ⇒ (1/2) 2x = (1/2) 6 ⇒ x = 3.
For example, x² + x – 6 = (x + 3)(x – 2). So, if we want to solve the quadratic equation x² + x – 6 = 0, we can rewrite it as (x + 3)(x – 2) = 0, which has two solutions: x = -3 and x = 2.
As previously, it’s a good idea to check these solutions to make sure they work and to catch any errors. In this case, (-3)² + (-3) – 6 = 9 – 3 – 6 = 0 and (2)² + 2 – 6 = 4 + 2 – 6 = 0, so the solutions x = -3 and x = 2 work.
The solutions are easy to see on a graph of the function y = x² + x – 6, to the right. The graph is a parabola opening upwards since the coefficient on x² is positive.
The solutions are the values of x when y = 0, where the graph crosses the x-axis in two places.
Two, one, or zero solutions
However, it’s not always the case that there are two solutions. For example:
- -x² – 2x = -x(x + 2).
Two solutions: x = -2, x = 0. - 4x² – 12x + 9 = (2x – 3)².
One solution: x = 3/2. - x² + 4x + 5 = (x² + 4x + 4) + 1 = (x + 2)² + 1.
No solutions.
More examples
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