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Math Help - Algebra prob

  1. #1
    Member ~berserk's Avatar
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    Algebra prob

    solve: 5-\sqrt{x+5}=2x+5
    i got 4x^2+x+5=0 for a quadratic

    but i believe it to be incorrect because the answer is x= -1 and i don't get that when i use the quad formula or complete the square so i think i messed up the quadratic
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  2. #2
    Junior Member
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    How about
    4x^2 - x - 5 = 0 ?
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  3. #3
    Member ~berserk's Avatar
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    why is it 4x^2-x-5=0 instead of 4x^2+x+5=0
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  4. #4
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    subtract 5 from both sides

    -sqr(x+5) = 2x

    square both sides

    x+5 = 4x^2

    take x + 5 from both sides

    0 = 4x^2 -x -5
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  5. #5
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by ~berserk View Post
    why is it 4x^2-x-5=0 instead of 4x^2+x+5=0
    5-\sqrt{x+5} = 2x+5

    -\sqrt{x+5} = 2x

    (\sqrt{x+5})^2=(-2x)^2

    x + 5 = 4x^2

    4x^2 - x -5 =0
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