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Math Help - Urgent, please help

  1. #1
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    Urgent, please help

    Hello all, thanks for the help.

    Ive been trying to figure this out most of the day and its driving me crazy. Its probably not that difficult for most of you so I apologize if it's too easy to post.

    solve for x:

    4+1/x-1/x^2=0

    thanks again.
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  2. #2
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    Re:

    Quote Originally Posted by Geomatt View Post
    Hello all, thanks for the help.

    Ive been trying to figure this out most of the day and its driving me crazy. Its probably not that difficult for most of you so I apologize if it's too easy to post.

    solve for x:

    4+1/x-1/x^2=0

    thanks again.
    RE:

    Before I help you out you are going to need to rewrite the problem because I can tell straight up that this isn't what you are trying to solve.

    qbkr21
    Attached Thumbnails Attached Thumbnails Urgent, please help-35.gif  
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  3. #3
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    Quote Originally Posted by qbkr21 View Post
    RE:

    Before I help you out you are going to need to rewrite the problem because I can tell straight up that this isn't what you are trying to solve.

    qbkr21
    ok , i apologize, im a newbie.

    lets try this again.

    4+(1/x)-(1/x^2)=0

    its 4 + the fraction of 1/x - the fraction of 1/x^2

    sorry for that.
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  4. #4
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    Re:

    Quote Originally Posted by Geomatt View Post
    ok , i apologize, im a newbie.

    lets try this again.

    4+(1/x)-(1/x^2)=0

    its 4 + the fraction of 1/x - the fraction of 1/x^2

    sorry for that.
    RE:
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  5. #5
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    Quote Originally Posted by qbkr21 View Post
    RE:
    Thank you so much for the help. Can I ask what program you use to write the problems out so I could maybe use something like that when I post again to avoid confusion?

    Also, you basically just cross-multiplied when you found the common denominator right?

    Thanks
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Geomatt View Post
    Thank you so much for the help. Can I ask what program you use to write the problems out so I could maybe use something like that when I post again to avoid confusion?
    he used a program called MathType

    but it may be easier for you to use LaTex, find the LaTex tutorial PDF here


    Also, you basically just cross-multiplied when you found the common denominator right?
    yes, that's what he did
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  7. #7
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    Quote Originally Posted by Geomatt View Post
    solve for x:

    4+\frac1x-\frac1{x^2}=0
    Another way could be

    4+\frac1x-\frac1{x^2}=0\iff{x^{-2}-x^{-1}-4=0}

    Now apply the quadratic formula and done.
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
    Another way could be

    4+\frac1x-\frac1{x^2}=0\iff{x^{-2}-x^{-1}-4=0}

    Now apply the quadratic formula and done.
    can we even apply the quadratic formula to something like that?

    i would have just multiplied through by -x^2, then we'd end up with the quadratic qbkr21 got right off the bat. then apply the quadratic formula
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  9. #9
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     <br />
\begin{aligned}<br />
x^{ - 2} + x^{ - 1} - 4 ~&=~ 0\\<br />
\left( {x^{ - 1} } \right)^2 + x^{ - 1} - 4 ~&=~ 0\\<br />
x^{ - 1} ~&=~ \frac{{ - 1 \pm \sqrt {17} }}<br />
{2}\\<br />
x_1 ~&=~ \frac{2}<br />
{{\sqrt {17} - 1}}\\<br />
x_2 ~&=~ - \frac{2}<br />
{{1 + \sqrt {17} }}<br />
\end{aligned}<br />
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Krizalid View Post
     <br />
\begin{aligned}<br />
x^{ - 2} + x^{ - 1} - 4 ~&=~ 0\\<br />
\left( {x^{ - 1} } \right)^2 + x^{ - 1} - 4 ~&=~ 0\\<br />
x^{ - 1} ~&=~ \frac{{ - 1 \pm \sqrt {17} }}<br />
{2}\\<br />
x_1 ~&=~ \frac{2}<br />
{{\sqrt {17} - 1}}\\<br />
x_2 ~&=~ - \frac{2}<br />
{{1 + \sqrt {17} }}<br />
\end{aligned}<br />
    ah! ok

    that's still a weird way to do it though
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  11. #11
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    I have a little typo, the second term is negative, but the idea is understandable.
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