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Math Help - quadratic formula

  1. #1
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    quadratic formula

    i'm trying to use the quadratic formula for 3x^2 - 12x + 5 = 0

    and got it to x = \frac {12 \pm \sqrt {-12^2 - 60}}{6} not sure if i'm going about it right so would like someone to point me in the right direction if possible

    thanks
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  2. #2
    Senior Member I-Think's Avatar
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    Yes you are very correct.
    But my teacher insisted on being picky so in that sentiment I'll just say that your notation, instead of

    Should be

    x=\frac{12\pm{\sqrt{(-12)^2-60}}}{6}

    The addition of brackets removes confusion, that's what my teacher used to say.
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  3. #3
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    ah thanks, the answer the book gives is x = 2 \pm \sqrt {\frac{7}{3}} so i can't see how you'd arrive at that from the answer i was at, could you show me the way to get to it please? there are also meant to be rackets under the root sign around the 7/3 like so: (7/3) but i don't know how to write it out using the brackets like that
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  4. #4
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    Quote Originally Posted by mark View Post
    i'm trying to use the quadratic formula for 3x^2 - 12x + 5 = 0

    and got it to x = \frac {12 \pm \sqrt {-12^2 - 60}}{6} not sure if i'm going about it right so would like someone to point me in the right direction if possible

    thanks
    a = 3

    b = -12

    c = 5

    x = \frac{12 \pm \sqrt{(-12)^2 - 4(3)(5)}}{2(3)}

    x = \frac{12 \pm \sqrt{144 - 60}}{6}

    x = \frac{12 \pm \sqrt{84}}{6}

    x = \frac{12 \pm \sqrt{4 \cdot 21}}{6}

    x = \frac{12 \pm 2\sqrt{21}}{6}

    x = \frac{6 \pm \sqrt{21}}{3} = 2 \pm \frac{\sqrt{21}}{3}

    note that \sqrt{\frac{7}{3}} = \frac{\sqrt{7}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{21}}{3}
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