algebra

sqrt(3x+4) + 4 = 2x

can you solve this with steps to help me please?

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  1. √(3x+4) + 4 = 2x
    √(3x+4) = 2x-4
    3x+4 = (2x-4)^2
    3x+4 = 4x^2 - 16x + 16
    4x^2 - 19x + 12 = 0
    (4x-3)(x-4) = 0
    x = 3/4 or 4

    But, squaring sometimes introduces extraneous roots. In this case, x = 3/4 does not work.

    √(3*3/4 + 4) + 4 = 2(3/4)
    √(9/4 + 4) + 4 = 3/2
    √(25/4) + 4 = 3/2
    5/2 + 4 = 3/2
    NO
    So, how did we get 3/4 as a solution?
    Because

    -√(25/4) + 4 = 3/2
    -5/2 + 4 = 3/2
    3/2 = 3/2

    when you square both sides, both the +√ and the -√ work. But the original equation only used the +√.

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