EQUATIONS CONTAINING VARIABLES UNDER ONE OR MORE RADICALS

Note:


Example 3:

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First make a note of the fact that you cannot take the square root of a negative number. Therefore, tex2html_wrap_inline97 .



Subtract 5 from both sides of the equation so that the radical term is isolated.

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Square both sides of the equation:

tex2html_wrap_inline99

tex2html_wrap_inline101




Subtract 8x and 5 from both sides of the equation.

tex2html_wrap_inline105

tex2html_wrap_inline107




Solve using the quadratic formula.

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Simplify.

tex2html_wrap_inline109

tex2html_wrap_inline111

tex2html_wrap_inline113




Check the solution x=3.605958 by substituting 3.605958 in the original equation. If after the substitution, the left side of the original equation equals the right side of the original equation, your answer is correct.

Check the solution tex2html_wrap_inline119 in the original equation. If after the substitution, the left side of the original equation equals the right side of the original equation, your answer is correct.

Since the left side of the original equation does not equal the right side of the original equation when 0.616264 is substituted for x, then 0.616264 is not 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. Since the only x-intercept is 3.605958, x=3.615958 and therefore is the only solution.


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.

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Author: Nancy Marcus

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