$x+ix=i$ solution? 3 years without math exercises , today I found a question in Quora about the solution of $x+ix =i$.
So I reasoned this way :
\begin{align} x+ix=i &\implies x=i−ix &\\
& \implies x=i(1−x) &\\
& \implies x/(1−x)=i\\
&\implies x^2/(1−x)^2=−1 \\
&\implies x^2=−[(1−x)^2]\\
& \implies x^2=−[1−2x+x^2] \\
&\implies x^2=−x^2+2x−1\\
&\implies 2x^2−2x+1=0\\
&\implies x^2−x+(1/2)=0
\end{align}
$$
a=1\quad b=−1 \quad c=1/2\\
\delta=b^2−4ac =−1=i^2
$$
so
$$x_1=(−b−i)/2∗a=(1−i)/2 \\
x_2=(−b+i)/2∗a=(1+i)/2$$
However the solution, as in site, is only $(1+i)/2$. What am I doing wrong? 
 A: Let's keep things simple. Suppose you are given $$x=1$$
Squaring both sides gives $$x^2=1$$
And the solutions of the quadratic equation $x^2=1$ are $x=1$ and $x=-1$. So the squared equation has more solutions that the original equation!
Whenever you square both sides of an equation like this, you must always go back to the original equation at the end, to check whether your solutions are still valid. In this case, you see that the solution $x=-1$ is not a valid solution of the equation $x=1$, so you reject it.
In your case, you must check both your solutions $x_1$ and $x_2$. You will find that only $x_2$ is a valid solution of the original equation. So you must discard $x_1$.
Having said that, there is no need to square both sides to solve your equation. Just write it as $$x(1+i)=i$$
and divide both sides by $1+i$.
A: $$x=\frac{i}{1+i}=\frac{i(1-i)}{(1+i)(1-i)}=\frac{i+1}{2}$$
A: Here’s what is wrong in your method. When you square both side by writing $\frac{x}{1−x}=i\Longrightarrow \frac{x²}{(1−x)²}=−1$ you have admitted others solutions which are not necessarily solution of the first equation. With this method you may verify which ,of the solution you have at the end, is a good one.
E.g
$$x=1\Longrightarrow x^2=1\Longrightarrow x=1, \text{ or } x=-1.$$ Thus we have to eliminate $-1$.
A: $[1].$ Suppose $A\implies B\implies C \implies x\in \{0,1\}.$ You cannot conclude, from this, that $x\in \{0,1\}\implies A.$ This does not mean that it isn't useful to use one-way implications. It's very useful but it's not the whole story. You have $A\implies x\in \{0,1\}$ and then  you may find 
(i).  $x=0\implies \neg A$ and $x=1\implies \neg A,$ and conclude that $A$ is false for all $x,$ OR
(ii). $x=0 \implies A$ and $x=1\implies  \neg A,$ and conclude that $A\iff x=0,$ OR
(iii). $x=0\implies \neg A$ and $x=1\implies A,$ and conclude that $A\iff x=1,$ OR
(iv) $x=0\implies A$ and $x=1\implies A,$ and conclude that $A\iff (x=0 \lor x=1).$
$[2].$ Let $a,b \in \Bbb R.$ Then $a+ib \ne 0$ iff $a,b$ are not both $0.$ And if  $a+ib\ne 0 $ then $a+ib$ has a multiplicative inverse $(a/r-ib/r)=(a-ib)/r$ where $r=a^2+b^2.$ Because $$(a+ib)(a-ib)/r=(a^2-i^2b^2)/r=(a^2+b^2)/r=1.$$ So if $x,y,z \in \Bbb C$ and $z\ne 0$ then $x=y\iff xz=yz.$ (Just as when $x,y,z\in \Bbb R$ and $z\ne 0$). 
If $a=b=1$ then $r= a^2+b^2=2$ so $1/(a+ib) =(a-ib)/r=(1-i)/2.$
Therefore $$x+ix= i\iff  x(1+i)=i\iff$$ $$\iff x\cdot (1+i)[1/(1+i)]=i\cdot[1/(1+i)] \iff$$ $$\iff x\cdot 1=i\cdot [(1-i)/2]\iff$$ $$\iff x=[i-i^2]/2=[i-(-1)]/2=(i+1)/2. $$
Notice the implications are two-way so there is nothing more to do.
In brief, solving a linear equation in one variable $(x)$ in $\Bbb C$ is done just the way you do it in $\Bbb R.$
