SOLVING QUADRATIC EQUATIONS

Note:




Solve for x in the following equation.

Example 4: text2html_wrap_inline253tex2html_wrap_inline324

Set the equation equal to zero by subtracting tex2html_wrap_inline326 and tex2html_wrap_inline328 from both sides of the equation.

eqnarray24

eqnarray31


eqnarray47

eqnarray59





Method 1:text2html_wrap_inline253Factoring

eqnarray68

eqnarray80

The left side of the equation is not easily factored, so we will not use this method.





Method 2:text2html_wrap_inline253Completing the square

Add tex2html_wrap_inline330 to both sides of the equation.

eqnarray99



Add tex2html_wrap_inline332 to both sides of the equation :

eqnarray120



Factor the left side and simplify the right side :

eqnarray132



Take the square root of both sides of the equation :

eqnarray141



Add tex2html_wrap_inline334 to both sides of the equation:

eqnarray150

eqnarray156

and

eqnarray164





Method 3:text2html_wrap_inline253Quadratic Formula

The quadratic formula is tex2html_wrap_inline336

In the equation tex2html_wrap_inline338 , a is the coefficient of the tex2html_wrap_inline340 term, b is the coefficient of the x term, and c is the constant. Simply insert 1 for a, tex2html_wrap_inline344 for b, and tex2html_wrap_inline346 for c in the quadratic formula and simplify

.

eqnarray195

eqnarray205

eqnarray217

eqnarray226

and

eqnarray234





Method 4:text2html_wrap_inline253Graphing

Graph tex2html_wrap_inline348 (the left side of the equation) and graph y= 0 (the right side of the equation). The graph of y=0 is nothing more than the x-axis. So what you will be looking for is where the graph of tex2html_wrap_inline348 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 located at 0.75 and -0.66667. This means that there are two real answers: x=0.75 and -0.66667.

The approximate answers are 0.75 and -0.66667.





Check these answers in the original equation.

Check the solution x=0.75 by substituting 0.75 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.

Since the left side of the original equation is equal to the right side of the original equation after we substitute the value 0.75 for x, then x=0.75 is a solution.


Check the solution x=-0.66667 by substituting -0.66667 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.

Since the left side of the original equation is equal to the right side of the original equation after we substitute the value -0.66667 for x, then x=-0.66667 is a solution.

The solutions to the equation tex2html_wrap_inline384 are -0.66667 and .75 .



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.

Author:Nancy Marcus

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