SOLVING QUADRATIC EQUATIONS

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
  3. Quadratic Formula
  4. Graphing

• All methods start with setting the equation equal to zero.

Solve for x in the following equation.

Example 1:

The equation is already set to zero.

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 2.

Method 1: Factoring

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

Method 2: Completing the square

Add 10 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 16 to both sides of the equation :

Method 3: Quadratic Formula

The quadratic formula is

In the equation ,a is the coefficient of the term, b is the coefficient of the x term, and c is the constant. Substitute 1 for a, -32 for b, and -10 for c in the quadratic formula and simplify.

Method 4: Graphing

Graph the left side of the equation, and graph the right side of the equation, The graph of is nothing more than the x-axis. So 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 32.309506 and one at -0.309506.

The answers are 32.309506 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=32.309506 by substituting 32.309506 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 32.309506 for x, then x=32.309506 is a solution.

Check the solution x=-0.309506 by substituting -0.309506 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 -0.309506 for x, then x= - 0.309506 is a solution.

The solutions to the equation are 32.309506 and - 0.309506.

Comment: You can use the exact solutions to factor the original equation.

Since

Since

The product

Since

then we could say

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

Multiply

Let's check to see if

The factors of are , and

If you would like to work another example, click on Example

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

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