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Math Help - [SOLVED] pythagoras

  1. #1
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    [SOLVED] pythagoras

    hi,

    having trouble seeing where i'm going wrong.

    c^2\=\a^2\+\b^2

    c\=\13,\a\=\3x,\b\=\3x

    13^2\=\(3x)^2\+\(2x)^2

    169\=\6x^2\+\4x^2

    169\=\10x^2

    16.9\=\x^2\

    x\=\pm\sqrt{16.9}


    x\=\sqrt{16.9}

    text book hast the answer as sprt{13}

    How?

    btw, I wrap math tags around these selected text and latex/syntex error comes up.

    anyone?
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  2. #2
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    Quote Originally Posted by nativecat View Post
    hi,

    having trouble seeing where i'm going wrong.

    c^2\=\a^2\+\b^2

    c\=\13,\a\=\3x,\b\=\3x

    13^2\=\(3x)^2\+\(2x)^2

    169\=\6x^2\+\4x^2 ... (3x)^2 = 9x^2, not 6x^2

    169\=\10x^2

    16.9\=\x^2\

    x\=\pm\sqrt{16.9}


    x\=\sqrt{16.9}

    text book hast the answer as sprt{13}

    How?

    btw, I wrap math tags around these selected text and latex/syntex error comes up. because the back slash {\} is only necessary for certain things.

    anyone?
    .
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  3. #3
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by nativecat View Post
    hi,

    having trouble seeing where i'm going wrong.

    c^2\=\a^2\+\b^2

    c\=\13,\a\=\3x,\b\=\3x

    13^2\=\(3x)^2\+\(2x)^2

    169\=\6x^2\+\4x^2

    169\=\10x^2

    16.9\=\x^2\

    x\=\pm\sqrt{16.9}


    x\=\sqrt{16.9}

    text book hast the answer as sprt{13}

    How?

    btw, I wrap math tags around these selected text and latex/syntex error comes up.

    anyone?
    The answers are x = \pm \sqrt{13} because you can have either a positive square root or a negative one. If you're talking about a length it will always be positive when dealing with scalars

    Skeeter pointed out that (3x)^2 = 3^2 \times x^2 = 9x^2
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