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Math Help - I hate radicals...

  1. #1
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    I hate radicals...

    i swear i'll never understand problems with radicals >.<

    anyway: \sqrt{\sqrt{x+8}+x}=2

    show work if you can..someday i'll understand those things -.-..always confudling me..
    Last edited by CaptainBlack; September 28th 2008 at 10:40 PM. Reason: language
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jcube69 View Post
    i swear ill nvr understand problems with gd radicals >.<

    anyways \sqrt{\sqrt{x+8}+x}=2

    show work if you can..someday ill understand those buggers -.-..always confudling me..
    square both sides, you get

    \sqrt{x + 8} + x = 4

    \Rightarrow \sqrt{x + 8} = 4 - x

    now square both sides again and continue

    (and please, do not make the mistake of thinking (4 - x)^2 = 4^2 - x^2, it is wrong and will tick your professor off)
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    square both sides, you get

    \sqrt{x + 8} + x = 4

    \Rightarrow \sqrt{x + 8} = 4 - x

    now square both sides again and continue
    i did that u get x + 8=x^2-8x+16then -8=x^2-9x or x^2-9x+8=0

    \frac{9\pm\sqrt{81-32}}{2} \frac{9\pm7}{2}=4.5\pm3.5 thus x=1,8?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jcube69 View Post
    i did that u get x + 8=x^2-8x+16then -8=x^2-9x or x^2-9x+8=0
    you're right so far

    \frac{9\pm\sqrt{81-32}}{2} \frac{9\pm7}{2}=4.5\pm3.5
    quadratic formula not needed. the quadratic was easily factorable

    thus x=1,8?
    yes, x = 8 or x = 1 are the solutions to the quadratic. are both solutions valid for our original problem though?

    also, that wasn't so hard, was it?
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  5. #5
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    yes, x = 8 or x = 1 are the solutions to the quadratic. are both solutions valid for our original problem though?

    also, that wasn't so hard, was it?
    bah didnt think about that..8 woulnt work because the sqrt of 12 is not 2

    thus it leaves only 1
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by jcube69 View Post
    bah didnt think about that..8 woulnt work because the sqrt of 12 is not 2

    thus it leaves only 1
    right. be aware that usually when you square things, you introduce erroneous solutions. you need to check when done
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  7. #7
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    Quote Originally Posted by Jhevon View Post
    right. be aware that usually when you square things, you introduce erroneous solutions. you need to check when done
    yup... tyvm ^.^
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