Note:

• A quadratic equation is a polynomial equation of degree 2.

• The ''U'' shaped graph of a quadratic is called a parabola.

• A quadratic equation has two solutions. Either two distinct real solutions, one double real solution or two imaginary solutions.

• There are several methods you can use to solve a quadratic equation:

1. Factoring
2. Completing the Square
4. Graphing

Solve for x in the following equation.

Problem 4.4a:

Solution:

Set the equation equal to zero by subtracting 2 from both sides.

If you have forgotten how to manipulate fractions, click on Fractions for a review.

Remove all the fractions by writing the equation in an equivalent form without fractional coefficients. In this problem, you can do it by multiplying both sides of the equation by 3.

Method 1:Factoring

The equation is not easily factored. Therefore, we will not use this method.

Method 2:Completing the square

Add 36 to both sides of the equation

Add to both sides of the equation:

Factor the left side and simplify the right side:

Take the square root of both sides of the equation:

Add 6 to both sides of the equation:

In the equation ,a is the coefficient of the term, b is the coefficient of the x term, and c is the constant. Substitute for a , for b, and for c in the quadratic formula and simplify.

Method 4:Graphing

Graph the equation, (formed by subtracting the right side of the equation from the left side of the equation). Graph (the x-axis). What you will be looking for is where the graph of crosses the x-axis. Another way of saying this is that the x-intercepts are the solutions to this equation.

You can see from the graph that there are two x-intercepts, one at 14.485281 and one at -2.485281.

The answers are and These answers may or may not be solutions to the original equations. You must verify that these answers are solutions.

Check these answers in the original equation.

Check the solution x=14.485281 by substituting 14.485281 in the original equation for x. If the left side of the equation

equals the right side of the equation after the substitution, you have found the correct answer.

• Left Side:

• Right Side:

Since the left side of the original equation is equal to the right side of the original equation after we substitute the value 14.485281 for x, then x=14.485281 is a solution.

Check the solution x=-2.485281 by substituting -2.485281 in the original equation for x. If the left side of the equation

equals the right side of the equation after the substitution, you have found the correct answer.

• Left Side:

• Right Side:

Since the left side of the original equation is equal to the right side of the original equation after we substitute the value -2.485281 for x, then x=-2.485281 is a solution.

The solutions to the equation are and

Comment:You can use the exact solutions to factor the left side of the original equation minus the right side of the original equation:

Since :

Since :

The product

Since and

then we could say

However the product of the first terms of the factors does not equal

Multiply by

Let s check to see if

The factors of are and

If you would like to review the solution to problem 4.4b, click on Problem

If you would like to test yourself by working some problems similar to this example, click on Problem

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Author: Nancy Marcus