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.

Example 4:

The equation is already set to zero.

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

Almost all precalculus students hate fractions before they renew their love for them. The following steps will transform the equation into an equivalent equation without fractions.

If you have forgotten what equivalent means, think of a dollar. You can represent the dollar with a dollar bill, 10 dimes, 20 nickels, or 100 pennies. All of these are equivalent because all have the value of a dollar. Got it. If not, click on Equivalence 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 462. All the denominators 6, 231, and 77 divide into 462 evenly.

How do we know this? and If we choose the set of factors

the set of factors of each denominator can be found in this set. This means that each of the denominators divides evenly into the product 462.

Method 1:Factoring

The left side of the equation can be factored as follows :

Method 2:Completing the square

Divide both sides of the equation by 77 .

Simplify:

Add 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 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 and one at .

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 by substituting 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 for x, then is a solution.

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

The solutions to the equation are and

Comment: Recall that when we solved this equation by factoring, we factored the expression not the original expression The product of the factors of does not equal the original expression because the product of the first terms of the factors must equal the first term of the original expression.

We need to add a constant factor.

Original Equation :

Product of factors equals What number do I multiply 77 by to get

Let's us see if equals the original We can do this using equivalence (evaluating both expressions by the same number). If the answers are the same, you have factored the original expression correctly. We can also this by algebraically by multiply the factors out and seeing if they equal the original expression

Equivalence:

Evaluate both expressions at x=2

This confirms the factors are correct using equivalence

Algebraically

Since the original expression is , we have shown the factors are correct algebraically.

We have illustrated that the solutions to the equation are and

By now you should be an expert on this type of problem. If you want a verification of this fact, test yourself over problems similar to the one above by clicking on Problem

[Algebra] [Trigonometry]
[Geometry] [Differential Equations]
[Calculus] [Complex Variables] [Matrix Algebra]

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Author: Nancy Marcus