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In this equation $$ix^{2}-2(i+1)x+(i-1)=0$$ one root is $i-1$ and another root is $-i$. How is it possible?

Solution I have tried :

Assume two roots are $A_1,A_2$ then $A_1+A_2=-b/a$ and $A_1A_2=c/a$. By using this formula I'm getting another root as $1/i$ not $-i$ but the correct answer is $-i$.

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    $\begingroup$ @SophieClad that's what how? $\endgroup$ – Mahender Reddy Yasa Mar 6 '15 at 6:44
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Very simple, you got $1/i$. See, $i^{-1}=-i$ because $i\cdot(-i)=1$, and if $xy=1$ then $x^{-1}=y$, here assume $y$ as $-i$

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$\frac {1}{i}$ = $-i$ .

To see this Multiply numerator and denominator of $\frac {1}{i}$ by $ i $

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    $\begingroup$ i know $i^2=-1$ but how it's possible that square of any number is equals to $-1$ $\endgroup$ – Mahender Reddy Yasa Mar 6 '15 at 6:53
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    $\begingroup$ Since $i\cdot(-i)$ equals $1$ that tells us $i^{-1}$ is $-i$. $\endgroup$ – anon Mar 6 '15 at 6:58
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    $\begingroup$ @anon How $i.(-i)=1$ tells $i^(-1) is -i$ $\endgroup$ – Mahender Reddy Yasa Mar 6 '15 at 7:05
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    $\begingroup$ $ab=1$ means $a^{-1}=b$. $\endgroup$ – anon Mar 6 '15 at 7:07
  • $\begingroup$ @SophieClad that is only true of quadratic equations with coefficients in R $\endgroup$ – ratchet freak Mar 6 '15 at 12:01

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