τρικλινο wrote:

Safal wrote:

2nd Solution is basically wrong. Why? Here is the explanation.

$x(x-5)=0$ Then there are two cases either $x=0$ or $x\neq 0$ .When we are admitting the case $x=0$ we cannot divide by $0$ . So, in the case we apply divison by $x$ then $x\neq 0$ is a solid prerequisite to do so.Thus, $x-5=0$ from $x(x-5)=0$ we must take the assumption in hand that $x\neq 0$ . For example take the extension of the same problem in $\mathbb{F}_5$ then the same problem reads $x^2=0$ ,implying only one solution with optimistic repetation of root $0$ , two times that is the multiplicity of $0$ in $x^2$ . Thankfully, we are lucky enough that we are in the field of $\text{char}$ $0$ .The reason is that, the book you mention was a book for below 10std (as far as I remember it is below 10th std) students, where prerequisite assumption is that ,we should work on field of $\text{char}$ $0$ .

so how do we get x=0 or x=5 ,since we assumed $x\neq 0$ .

If you read carefully I haven't said that we cannot get $x=0$ .The assumption whenever $x\neq 0$ we get $x=5$ else we get the case $x=0$ .I can explain you more but the fact is I cannot use argument of field theory to explain it in total details.The reason why it's actually the case lies in field theory logics.