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Thread: Can I get help with this triq problem?

  1. #1
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    Can I get help with this triq problem?

    I'm trying to find X P( x, -2/7)

    $\displaystyle x^2+ -2/7=1$ then I get
    $\displaystyle x^2+ (-2/7)^2=1$
    Than I get $\displaystyle x^2=1-4/49= $ $\displaystyle x^2= \frac{49} {49}- \frac{4}{49}=\frac{45} {49}$ Then you get $\displaystyle x=\sqrt \frac{45} {49}$The answer is suppose to be $\displaystyle x=+ or -\sqrt{5}$.

    How do you get $\displaystyle x= + - \sqrt{5}$

    Thanks

    Jason
    Last edited by Darkhrse99; Jun 9th 2008 at 05:08 PM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Darkhrse99 View Post
    I'm trying to find X P( x, -2/7)

    $\displaystyle x^2+ -2/7=1$ then I get
    $\displaystyle x^2+ (-2/7)^2=1$
    Than I get [tex]x^2=1-4/49= $\displaystyle x^2= 49/49-4/49=x^2=45/49$ The answer is suppose to be[tex] x=\sqrt{5}.

    How do you get sqrt + - 5?

    Thanks

    Jason
    Can I just be the first to say

    Huh?

    You started with
    $\displaystyle x^2 \pm \frac{2}{7} = 1$

    and went to
    $\displaystyle x^2 + \left ( -\frac{2}{7} \right )^2 = 1$

    How did you go from the first to the second step? The first does not imply the second.

    What exactly is it you are trying to find? "X P (x, -2/7)" makes no sense to me.

    -Dan
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  3. #3
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    Sorry for the confusion, the math code is kicking my butt.

    I'm trying to find point X on the unit circle.

    The book gave me the Y cord of $\displaystyle \frac {-2} {7}$ . I need to find X.

    The formula $\displaystyle x^2+ y^2 =1$ is used to find the answer.

    I plugged in $\displaystyle x^2 + \frac {-2} {7}squared=1$

    Than I get $\displaystyle x^2=1-\frac{4} {49} next I get$ $\displaystyle x^2= \frac{49} {49}- \frac{4}{49}=\frac{45} {49}$ Then you get $\displaystyle x=\sqrt \frac{45} {49}$The answer is suppose to be $\displaystyle x=+ or -\sqrt{5}$.
    Last edited by Darkhrse99; Jun 9th 2008 at 05:49 PM.
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  4. #4
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    I found the answer. The book was wrong. The answer is $\displaystyle 3\sqrt{5}/7 $
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Darkhrse99 View Post
    Sorry for the confusion, the math code is kicking my butt.

    I'm trying to find point X on the unit circle.

    The book gave me the Y cord of $\displaystyle \frac {-2} {7}$ . I need to find X.

    The formula $\displaystyle x^2+ y^2 =1$ is used to find the answer.

    I plugged in $\displaystyle x^2 + \frac {-2} {7}squared=1$

    Than I get $\displaystyle x^2=1-\frac{4} {49} next I get$ $\displaystyle x^2= \frac{49} {49}- \frac{4}{49}=\frac{45} {49}$ Then you get $\displaystyle x=\sqrt \frac{45} {49}$The answer is suppose to be $\displaystyle x=+ or -\sqrt{5}$.
    The solution is
    $\displaystyle \pm \frac{3 \sqrt{5}}{7}$
    so your given answer was wrong. You did it (mostly) right.

    -Dan
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  6. #6
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    Yes, mostly right. I was stumped all day. The solutions manual said $\displaystyle \pm\sqrt{5} $ but the text book said $\displaystyle
    \pm \frac{3 \sqrt{5}}{7}
    $
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