SOLVING QUADRATIC EQUATIONS WITH RADICAL COEFFICIENTS

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 2:

The equation is already set to zero.

Simplify the equation.

can be rewritten as:

Method 1:Factoring

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

Method 2:Completing the square

Add 8 to both sides of the equation .

Divide both sides of the equation by and simplify.

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:

Subtract from both sides of the equation:

and

The exact answers are and the approximate answers are

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 for a, for b, and -8 for c in the quadratic formula and simplify.

The exact answers are and the approximate answers are

Method 4:Graphing

Graph the left side of the equation, Graph y=0 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 0.77292396 and one at -2.439590637.

The answers are 0.77292396 and -2.43950637. 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=0.77292396 by substituting 0.77292396 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 0.77292396 for x, then x=0.77292396 is a solution.

Check the solution x=-2.439590637 by substituting -2.439590637 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 -2.439590637 for x, then x=-2.439590637 is a solution.

The exact answers are and the approximate answers are

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

If you would like to go back to the equation table of contents, click on Contents

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

S.O.S MATHematics home page

Do you need more help? Please post your question on our S.O.S. Mathematics CyberBoard.

Copyright © 1999-2024 MathMedics, LLC. All rights reserved.
Contact us
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA
users online during the last hour