Quadratic Equations
The general form of quadratic equations (or a second-degree equation) is
For example,
is a quadratic equation. In this example,
Solving Quadratic Equations by Factoring
Factoring is a powerful technique for solving quadratic equations. A quadratic equation in the form
where
Thus, the equation is satisfied when either
Factoring a Simple Quadratic
Solve the quadratic equation
Solution
To solve the equationFactoring a Quadratic with a Leading Coefficient
Solve the quadratic equation
Solution
To solve this equation by factoring, we write:Completing the Square
If we can rewrite the quadratic equation
then we can easily solve it by taking the square root of each side (see Section: Power Equations). The left hand side is a perfect square---the square of
So in a perfect square, the constant term is the square of half of the coefficient of
A binomial
This method which is called completing the square has many applications in different parts of mathematics. One of these applications is in solving the quadratic equations and deriving the quadratic formula we have seen in this section.
To solve
we add
Therefore, the sum of the first three terms makes a perfect square
For example, let us solve
First, transpose 5; that is, add 5 to both sides
If now we add
The solutions therefore are
- If there is a constant
multiplying the term, then we first factor out that constant and then complete the square, as illustrated in the following example.
Solve
Solution
Quadratic Formula
The solutions (or roots) of the quadratic equation (i) can always be found using the following general formula, called the quadratic formula:
Proof
The quadratic formula can be derived using completing the square. Here's a step-by-step proof: 1. **Start with the General Form**:When we apply the general formula to the equation
The part of the quadratic formula that is under the square root,
- Factorization: When Equation
has two real solutions and , it factors as For example, because the solutions of are 1 and , we have
Factor
Solution
It may not be easy to factor it by trial and error. So we find the solutions of- Sum and product of roots: If
and are the solutions of then Because if we expand , we have So by comparing with , we realize that and .
For example, the solutions of
- If
, then one of the roots is 1, because if we substitute 1 for , we get - If
, then one of the roots is , because if we substitute for we get