# How is another root -i?

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 $-i$.

• @SophieClad that's what how? – Mahender Reddy Yasa Mar 6 '15 at 6:44

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 $-i$
$\frac {1}{i}$ = $-i$ .
To see this Multiply numerator and denominator of $\frac {1}{i}$ by $i$
• i know $i^2=-1$ but how it's possible that square of any number is equals to $-1$ – Mahender Reddy Yasa Mar 6 '15 at 6:53
• Since $i\cdot(-i)$ equals $1$ that tells us $i^{-1}$ is $-i$. – anon Mar 6 '15 at 6:58
• @anon How $i.(-i)=1$ tells $i^(-1) is -i$ – Mahender Reddy Yasa Mar 6 '15 at 7:05
• $ab=1$ means $a^{-1}=b$. – anon Mar 6 '15 at 7:07