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 4.5b:

are the exact answers using the Completing the Square Method.

Even though the two answers look different, they are equivalent because both yield the same approximate answers of and

Solution:

Simplify the equation by simplifying the radicals.

Eliminate the denominator 5 by multiplying both sides by 5.

Method 1:Factoring

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

Method 2:Completing the square

Subtract 1 to both sides of the equation

Divide both sides by

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 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, (the left side of the original 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 -0.02354671551 and one at -1.20119815588.

The answers are -0.02354671551 and -1.20119815588. 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=-0.02354671551 by substituting -0.02354671551 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 -0.02354671551 for x, then x=-0.02354671551 is a solution.

Check the solution x=-1.20119815588 by substituting -1.20119815588 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 -1.20119815588 for x, then x=-1.20119815588 is a solution.

The solutions to the equation are-0.02354671551 and-1.20119815588.

If you would like to review the solution to problem 4.5c, click on Problem

If you would like to go back to the beginning of the quadratic section, click on Quadratic

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Author: Nancy Marcus