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$-\frac{1}{2}z^{-2}+\frac{1}{2}z^{-1}+1=0$

For $\frac{1}{2}x^{2}+\frac{1}{2}x^{1}+1=0$

We could use the pq-Formula. I guess it is possible to use this here, but how? If not, what techniques are used for negative exponents?

Tried use factorization:

$-\frac{1}{2}z^{-2}+\frac{1}{2}z^{-1}+1=0$

$-\frac{1}{2}z^{-2}+\frac{1}{2}z^{-1}+1=0$

$=1+\frac{1}{2}(-z^{-2}+z^{-1})$ $\equiv -1 = \frac{1}{2}(-z^{-2}+z^{-1})$

$-z^{-2}+z^{-1}=0$ Here I would be stuck with

$z^{-2}=z^{-1}$

I am trying to improve my math and questions here, if there is anything wrong, just tell me.

Also I would like to know, if I could just multiply by z.

For negative exponents, this would make the problem easier.

For positive exponents, this would make the problem harder.

Is this method legit?

Edit:

Try to multiply by z until all exponents are positive lead me to:

$z^2+\frac{1}{2}z-\frac{1}{2}=0$

$p_{1,2}= -\frac{1}{4} \pm \sqrt{ \frac{7}{16} }$

For the squareroot of minus values imaginary numbers are required, right?

Thanks for the comments, will try what you commented.

Rapiz
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    make the substitution $u = 1/z$. Solve the resulting quadratic for $u$. Then back-substitute to find $z$. – DOUGLAS BRUNSON Mar 05 '22 at 19:03
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    Note that $z$ can't be $0$. So multiply the equation by $z^2$, and you get a simple quadratic. – Andrei Mar 05 '22 at 19:04
  • Note that when you went from the first equation to the second, you did a substitution $z\to z^{-1}$. For clarity, use a different variable. Then equations 3 and 4 are the same. Equation 5 you somehow transformed $z^{-1}$ to $z$. – Andrei Mar 05 '22 at 19:07
  • For the second equation, it is no transform. Just the situation without negative exponents, where I would know, what to do. Will change the variable, thanks. – Rapiz Mar 05 '22 at 19:11
  • Basically I've tried what you suggested and got to the square root of a negative value. Did it with multiplaction by z until the exponents are all positve. @DOUGLASBRUNSON basically the same like u=1/z, thus this should get me to the same point? Is my pq formula wrong? – Rapiz Mar 05 '22 at 19:18
  • I think the -1/2 at the end is an error. Check your work and confirm. – DOUGLAS BRUNSON Mar 05 '22 at 19:23
  • Well the substitution seems to get me an easier problem. Thanks! Thanks @DOUGLASBRUNSON, will check it! – Rapiz Mar 05 '22 at 19:23
  • NVM. The question title and the first line have an inconsistent sign. I agree with your last line. Square roots of negative numbers are fine. Just means its a complex root. – DOUGLAS BRUNSON Mar 05 '22 at 19:25
  • Also, I did some major rhymes in my previous comment, which was actually unintentional. – DOUGLAS BRUNSON Mar 05 '22 at 19:27
  • Found the problem with the square root, not at least the previous way should be correct! On my tab I wrote 1/16 - 1/2 instead of 1/16 + 1/2. – Rapiz Mar 05 '22 at 19:28
  • @DOUGLASBRUNSON nice verse. Math Math City, 10 times 5 is 50 – Rapiz Mar 05 '22 at 19:48

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