given the equation Square root of x minus 5 + 7 = 11 Solve for x and identify if it is an extraneous solution.

Question

given the equation Square root of x minus 5 + 7 = 11 Solve for x and identify if it is an extraneous solution.

in progress 0
2022-06-23T09:35:52+00:00 2 Answers 0

Answers ( 2 )

  1. Anna
    0
    2022-06-23T09:37:38+00:00

    Answer:

    x=21 and solution is not extraneous.

    Step-by-step explanation:

    Given : Expression [tex]sqrt{(x-5)}+7=11[/tex]

    To find : The solution for x and identify if it is an extraneous solution?

    Solution :

    Expression [tex]sqrt{(x-5)}+7=11[/tex]

    [tex]sqrt{(x-5)}=11-7[/tex]

    [tex]sqrt{(x-5)}=4[/tex]

    Taking square both side,

    [tex](sqrt{(x-5)})^2=4^2[/tex]

    [tex]x-5=16[/tex]

    [tex]x=21[/tex]

    Substituting the value back in the equation to find this is an extraneous solution or not.

    [tex]sqrt{(21-5)}+7=11[/tex]

    [tex]sqrt{(16)}+7=11[/tex]

    [tex]4+7=11[/tex]

    [tex]11=11[/tex]

    This is true.

    Which means solution is not an extraneous solution.

  2. Kylie97
    0
    2022-06-23T09:37:45+00:00

    sqrt(x) – 5+7 =11
    combine like terms
    -5 +7 = 2

    sqrt(x) + 2 = 11
    subtract 2 from each side

    sqrt(x) = 9

    square both sides

    x = 9^2

    x = 81

    an extraneous solution would be an invalid solution

    in this case it is not an extraneous solution

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