Hurry 50 points for correct answer Solve the equation. Identify any extraneous situations. x=√6x+7 1 is a solution t

Question

Hurry 50 points for correct answer
Solve the equation. Identify any extraneous situations.
x=√6x+7

1 is a solution to the original equation. The value –7 is an extraneous solution.

7 is a solution to the original equation. The value –1 is an extraneous solution.

7 and –1 are solutions.

7 and 1 are both extraneous solutions.

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Samantha 2 hours 2021-10-14T13:50:34+00:00 2 Answers 0

Answers ( )

    0
    2021-10-14T13:52:05+00:00

    Answer: “7 is a solution to the original equation. The value –1 is an extraneous solution.”

    Step-by-step explanation:

    0
    2021-10-14T13:52:19+00:00

    Answer: “7 is a solution to the original equation. The value –1 is an extraneous solution.”

    Step-by-step explanation:

    The equation x=\sqrt{6x+7} can be solved by squaring both sides:

    x^2 = 6x + 7\\\\x^2 - 6x - 7 = 0\\\\(x-7)(x+1) = 0

    We can see that -1 and 7 are solutions, but make sure they are not extraneous by substituting them in the original equation:

    1 = \sqrt{-1}\\ 7 = \sqrt{49}

    The square root of 49 equals 7, but the square root of -1 is an imaginary number.

    The correct choice is “7 is a solution to the original equation. The value –1 is an extraneous solution.”

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45:7+7-4:2-5:5*4+35:2 =? ( )