In standard form, quadratic equations take the form and can be solved in various ways. In general, they are solved for the value(s) of that make the equation equal to Thus, becomes Graphically, all points with a -coordinate of are the -intercepts of the function — or the zeros of the parabola. That means, solving a quadratic equation leads to finding the zeros of the parabola. Since a parabola can have or zeros, a quadratic equation can have or solutions.
If neither side of the equation equals rearrange it so that all terms are on the same side. If there is more than one term of the same degree, simplify the equation by combining like terms.
The side with the variables can now be seen as a function, Plot that function in a coordinate plane. For the equation the function is
Now, find all the points on the graph that have the -coordinate
The -coordinates of any identified points solve the equation,
In the example, the -coordinates are and Note that solving an equation graphically does not necessarily lead to an exact answer. To verify a solution, substitute it into and evaluate the expression. Verifying the solution is done by evaluating
Simple quadratic equations take the form and are solved using inverse operations. Once remains on the left-hand side, the equation can be written as where . The value of gives the number of solutions the equation has.
seeits factors. This expression can be factored by finding a pair of integers whose product is which here is and whose sum is which in the example is
The first step is to find all possible pairs of integers that multiply to In this case, and Thus, their product is To find all factor pairs, start with the pair where one factor is The other factor must then be Then continue with the pair where one of the factors is and so forth. In this case, there are three pairs.
If the given expression is factorable, one of the factor pairs will add to equal In this case, Here, and is the only factor pair that adds to
Now, use the factor pair to rewrite the -term of the original expression as a sum. Since the factor pair is and the middle term can be written as This gives the following equivalent expression. Notice that the expression hasn't been changed. Rather, it has been rewritten.
Now, the expression has four terms, which can be grouped into the first two terms and the last two terms. Then, the GCF of each group can be factored out. Here, begin with the first group of terms, and Notice how each can be written as a product of its factors. The GCF is Therefore, it's possible to factor out
Next, repeat the same process with the last two terms. In this case, and do not have any common factors, but it's always possible to write expressions as a product of and itself.
If all previous steps have been performed correctly, there should now be two terms with a common factor, which can be seen as a repeated parentheses. Factoring out of both terms gives This means that can be written in factored form as
Solve the following equation by factoring.