EQUATIONS CONTAINING VARIABLES UNDER ONE OR MORE RADICALS

Note:


  • Problem 2.3e:
    tex2html_wrap_inline100

    Answer: x=3.46939

    Solution:

    First make a note of the fact that you cannot take the square root of a negative number. Therefore, tex2html_wrap_inline104 .



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

    displaymath74




    Square both sides of the equation and simplify:

    tex2html_wrap_inline106

    tex2html_wrap_inline108

    tex2html_wrap_inline110




    Subtract 100x+25 from both sides of the equation.

    displaymath75




    Solve using factoring:

    or

    tex2html_wrap_inline122

    tex2html_wrap_inline124



    The answers are tex2html_wrap_inline126




    Check the solution x=0 by substituting 0 for x in the original equation. If after the substitution, the left side of the original equation equals the right side of the original equation, 0 is a solution.

    Since the left side of the original equation does not equal the right side of the original equation after 0 was substituted for x, x=0 is not a valid solution.



    Check the solution x=3.46939 by substituting 3.46939 for x in the original equation. If after the substitution, the left side of the original equation equals the right side of the original equation, 3.46939 is a solution.

    Since the left side of the original equation equals the right side of the original equation after 3.46939 was substituted for x, then x=3.46939 is a valid solution.



    You can check the answer by graphing the equation:

    displaymath76

    The graph represents the right side of the original equation minus the left side of the original equation. You can see that there is one x-intercept, at x=3.46939. This means that there is one solution x=3.45939.


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    Author: Nancy Marcus

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