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Math Help - Should be a simple solve for X

  1. #1
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    Should be a simple solve for X

    Hi, I've run into another problem on my algebra refresher. I must be doing something wrong here but I just can't see it.

    \sqrt{10-x}+x = 4

    Need to solve for X, of course I know the answer should be 1 but I can't seem to arrive at it.

    My steps:
    1: Put the lone x onto the right side.

    \sqrt{10-x} = 4-x

    2:Square both sides (to get rid of the square root on left side).

    10-x = 16+x^2

    3: Rearrange.
    0 = x^2 + x + 6

    4: At this point I could try putting this into the quadratic formula but it has a negative square root.

    A mistake I may have made is in step 2, was I wrong to make the x^2 on the right hand side a positive? But even if it is left as a negative my answer comes out as...

    x^2 - x - 6

    Which I can make into (x-3)(x+2) I think, which would make x = 3, -2 but neither of those are the right answer of 1.

    Am I doing this problem very wrong? Thanks in advance.
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  2. #2
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    (4 - x)^2 \neq 16 + x^2.


    Remember that (a + b)^2 = (a + b)(a + b) = a^2 + 2ab + b^2.


    So that means (4 - x)^2 = 16 - 8x + x^2.
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  3. #3
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    OH! I always made that same mistake when I was in high school too.

    Thanks for your help, I've been staring at this for a while now.
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