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.

2. ## Re:

Originally Posted by Geomatt
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

3. Originally Posted by qbkr21
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.

4. ## Re:

Originally Posted by Geomatt
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:

5. Originally Posted by qbkr21
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

6. Originally Posted by Geomatt
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

7. Originally Posted by Geomatt
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.

8. Originally Posted by Krizalid
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

9. 
\begin{aligned}
x^{ - 2} + x^{ - 1} - 4 ~&=~ 0\\
\left( {x^{ - 1} } \right)^2 + x^{ - 1} - 4 ~&=~ 0\\
x^{ - 1} ~&=~ \frac{{ - 1 \pm \sqrt {17} }}
{2}\\
x_1 ~&=~ \frac{2}
{{\sqrt {17} - 1}}\\
x_2 ~&=~ - \frac{2}
{{1 + \sqrt {17} }}
\end{aligned}

10. Originally Posted by Krizalid

\begin{aligned}
x^{ - 2} + x^{ - 1} - 4 ~&=~ 0\\
\left( {x^{ - 1} } \right)^2 + x^{ - 1} - 4 ~&=~ 0\\
x^{ - 1} ~&=~ \frac{{ - 1 \pm \sqrt {17} }}
{2}\\
x_1 ~&=~ \frac{2}
{{\sqrt {17} - 1}}\\
x_2 ~&=~ - \frac{2}
{{1 + \sqrt {17} }}
\end{aligned}
ah! ok

that's still a weird way to do it though

11. I have a little typo, the second term is negative, but the idea is understandable.