Is $0$ the root of the equation $\frac{x^2}{ x}= 0$? I want to understand if I understand the concept of equation correctly. For this I want to know if $0$ is the root of the equation $\frac{x^2}{x} = 0$. If we simplify the equation then we can get equation $x = 0$ and then we have solution $0$. But if we replace $x$ by $0$ in the original equation then we get expression $\frac{0^2}{0} = 0$ which doesn't make sense. I'm interested in your opinion.
 A: No, this system is indeterminate and there is no solution for $x=0$. The key point in this question is to figure out how simplifying a system can often hide certain properties of it, or how simplifications can be incorrectly made by violating certain mathematical conditions.
Take for instance, the equation you have:
$$\frac{x^2}{x} = 0$$
All mathematical systems are dependent on how they are simplified. In this form, there can be no solution, because have the form $\frac{0^2}{0}$ which is indefinite.
The problem with the method of simply bringing the denominator over to the right-hand side (RHS) is that it simplifies the equation and loses the importance of the fraction representation. The indefinite form specifically has a condition that if it is of the form $\frac{0}{0}$, then it cannot be simplified by shifting the denominator to the opposite side (LHS or RHS). Thus, when simplifying the system, you missed this important condition.
Bottom line, canceling 2 terms on the RHS and LHS require the term to be divided across both sides, but this yields an indefinite answer if the division creates an indeterminate/indefinite form.
Here's a famous example of a proof that goes wrong because of the incorrect application of the above condition:
$$
\begin{align}
Let, \ a & = b \\
a^2 & = ab \\
a^2 + a^2 & = a^2 + ab \\
2a^2 & = a^2 + ab \\
2a^2 - 2ab & = a^2 + ab - 2ab \\
2(a^2 - ab) & = a^2 - ab \\
2 & = 1
\end{align}
$$
The error lies in the last step, since you cannot cancel out the term $a^2-ab$ from LHS & the RHS. The reason is because when you cancel out the term, you are essentially dividing the term across both sides. However, you know that $a^2 - ab = 0$ from the assumption $a=b$, meaning that your equation becomes:
$$
\begin{align}
2\left[\frac{a^2-ab}{a^2-ab}\right] & = \left[\frac{a^2-ab}{a^2-ab}\right] \\
2\left[\frac{0}{0}\right] & = \left[\frac{0}{0}\right]
\end{align}
$$
Which is an indeterminate/indefinite form, rendering the proof incorrect. The same logic extends to your question.
A: No, the equation has no solutions. The problem comes from the step of simplifying the equation. In this step you are dividing by $x$, but in that case you are dividing by $0$, which is not allowed. If you assume that $x\not=0$, you can simplify and get $x=0$, which is a contradiction. So the equation has no solutions
