Patrick solved the rational equation $$ \frac{3x(x+2)}{x^2-4}=0$$ as follows:
(1) He eliminated the fraction by multiplying both sides of the equation by $(x^2-4 )$ and got the equation: $$ 3x(x+2)=0 $$
(2) The product is equal to zero if one of the factors is zero, i.e.: $$3x=0 \mathrm{~or~}(x+2)=0$$
(3) By solving the above equations, he obtained two solutions: $$ x=0 \mathrm{~or~} x=−2 $$
(4) He checked the first solution by substituting $x=0$ into the equation: $$ L=\frac{3\cdot 0 \cdot (0+2)}{0^2-4}=\frac{0}{-4}=0\Rightarrow L=R $$
(5) He checked the second solution by substituting $x=-2$ into the equation: $$ L=\frac{3\cdot (-2)\cdot (-2+2)}{-2^2-4}=\frac{-6\cdot 0}{-8}=\frac{0}{-8}=0 \Rightarrow L=R $$
The classmates made comments on Patrick's solution. Which one is wrong?
Henry says that by multiplying both sides of the equation by $(x^2-4 )$, he lost one solution, $x=2$.
John claims that by multiplying both sides of the equation by $(x^2-4 )$, he obtained the incorrect solution $x=-2$.
Erika says that if Patrick had determined the condition $x^2−4\neq 0$ right at the beginning, there would be no need for the checks.
Paul claims that Patrick made an error in step (5).
Sarah says that if we want to perform an equivalent transformation of an equation, we can only multiply the equation by a non-zero expression.
The given equation has only one solution, $x=0$.