EQUATIONS CONTAINING VARIABLES UNDER ONE OR MORE RADICALS

Recall the following:


Solve for x in the following equation.

Problem 2.5b:

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Answer: x=5.

Solution:

First make a note of the fact that you cannot take the square root of a negative number. Therefore, the tex2html_wrap_inline134 term is valid only if tex2html_wrap_inline136 , the term tex2html_wrap_inline138 is valid only if tex2html_wrap_inline140 , and the term is valid only if tex2html_wrap_inline144 . The equation is valid if all three terms are valid, therefore the domain is restricted to the common domain of the three terms or the set of real numbers tex2html_wrap_inline146




Square both sides of the equation and simplify.

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Isolate the tex2html_wrap_inline148 term and simplify.

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Square both sides of the equation and simplify.

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Use the quadratic formula to solve for x.

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Discard the second answer because 0.391676 is not in the domain.




The answers is x=5.




Check the solution by substituting 5 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 5 for x, then x=5 is a solution.




You can also check the answer by graphing the equation:

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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 one x-intercept at 5, then the only solution is x=5.


If you would like to review the solution to problem 2.5c, click on solution.

If you would like to go back to the problem page, click on problem.

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

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Author:Nancy Marcus

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