Note:
Example 1:
First make a note of the fact that you cannot take the square root of
a negative number. Therfore, the term is valid only if
, the second term
is valid only if , and the term
is valid only if . The
equation is valid if all the three terms are valid, therefore the
domain is restricted to the common domain of the three terms or the set
of real numbers .
Square both sides of the equation and simplify.
Isolate the term.
Square both sides of the equation and simplify.
Use the quadratic formula to solve for x.
The answers are .
Check the solution by substituting 1 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 equals the right side of the original equation after we substituted our solution for x, then x=1 is a solution.
Check the solution by substituting -4.244444 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 does not equal the right side of the original equation after we substituted -4.244444 for x, then the solution x= -4.244444 is not valid and not a solution after all.
You can also check the answer by graphing the equation:
The graph represents the right side of the original equation minus the left side of the original equation. The x-intercept(s) of this graph is (are) the solution(s). Since there is just on x-intercept at 1, then the only solution is x=1.
If you would like to test yourself by working some problems similar to this example, click on Problem.p> If you would like to go back to the equation table of contents, click on Contents.
Do you need more help? Please post your question on our S.O.S. Mathematics CyberBoard.
Author:Nancy Marcus