Show that the statement $p:$ 'If $x$ is a real number such that $x^{3}+4x=0$,then $x$ is $0$' is true by the method of contradiction.

(N/A) Let $p$ be the statement: 'If $x$ is a real number such that $x^{3}+4x=0$,then $x$ is $0$.'
To prove $p$ by contradiction,we assume that $p$ is false.
The negation of the statement 'If $q$,then $r$' is '$q$ and not $r$'.
Thus,we assume that $x$ is a real number such that $x^{3}+4x=0$ and $x \neq 0$.
Given $x^{3}+4x=0$,we can factor this as $x(x^{2}+4)=0$.
This implies $x=0$ or $x^{2}+4=0$.
Since we assumed $x \neq 0$,we must have $x^{2}+4=0$,which means $x^{2}=-4$.
However,for any real number $x$,$x^{2} \geq 0$.
Therefore,$x^{2}=-4$ has no real solution.
This contradicts our assumption that $x$ is a real number.
Hence,our assumption that $p$ is false must be incorrect.
Therefore,the statement $p$ is true.