Why is $x=2 \implies (x-2)(x-3)=0$ false? Let $P(x)$ be the equation $x=2$
and $Q(x)$ be the equation  $(x-2)(x-3)=0$
By definition of implication I see that $P(x)$ implies $Q(x)$...
As I see it, any premise that is false can give any consequence.
With $x=2$ both sides of the arrow are true.. so implication is true.
It is clear that $Q(x)$ implies $P(x)$ and it is also clear that $P(x)$ and $Q(x)$ are not equivalent. Since $Q(x)$ implies $P(x)$ and $P(x)$ and $Q(x)$ are not equivalent it follows that $P(x)$ cannot imply $P(x)$.
Where am I wrong?
EDIT: I knew that $P(x)\Leftrightarrow Q(x)$ was not valid. And I thought wrongly that $P(x)\Leftarrow Q(x)$ is valid which should, if it was valid, imply that $P(x)\nRightarrow Q(x)$. But testing $P(x)\Rightarrow Q(x)$ by the truth table showed $P(x)\Rightarrow Q(x)$ is valid which confused me. I was simply wrongly thinking $P(x)\Leftarrow Q(x)$ which is the correct answer. I have also posted a soultion below, which typo is corrected.
 A: $$(x-2)(x-3)=0 \iff x = 2 \ \, \text{or} \, \ x = 3$$
If $x=2$, then the hypothesis of the RHS is satisfied and hence we have the LHS.
I don't see why that is supposed to be false.
A: Perhaps you mean the converse $Q(x) \Rightarrow P(x)$...that is false. If $(x-2)(x-3) =0$, then $x=2$ or $x=3$, so it is not necessarily the case that $x=2$.
A: On your comment to Jack Bauer's answer you write 

"it is supposed to be false since x=3 is a solution... but by the definition of implication I see it true... what is wrong?"

The problem is when you say "It is supposed to be false since $x=3$ is a solution." The fact that $x=3$ is a solution does not make "$P\implies Q$" false; it makes "$Q\implies P$" false. There is absolutely no problem with saying "$P$ implies $Q$, but is not equivalent to $Q$."
A: I think the problem is in how to understand logical expressions and how to use them. If you look at the truth table of $P\Rightarrow Q$, you have:
\begin{array}{ccc}
P&Q&(P\Rightarrow Q)\\\hline
T&T&T\\
T&F&F\\
F&T&T\\
F&F&T
\end{array}
Therefore you are correct in saying that if $P$ is false, then $P\Rightarrow Q$ is true, no matter what $Q$ is. This doesn't mean, however that $Q$ must be true. 
The way to use the logical statements is:
\begin{align}
P \text{ is true}&&\color{blue}{\text{and}}&&(P\Rightarrow Q)\text{ is true}
\end{align}
imply that
\begin{equation}
Q\text{ is true}.
\end{equation}
Let's apply this to $Q\Rightarrow P$. In order to infer that $P$ is true, you need
\begin{align}
Q \text{ is true}&&\color{blue}{\text{and}}&&(Q\Rightarrow P)\text{ is true}.
\end{align}
Here we run into a problem. As was mentioned in other answers and comments, if $Q$ is true, this means that $x=2$ or $x=3$. So $Q\Rightarrow P$ is false. We can't infer that $P$ is true.
A: There is a tacit "$\forall x$" here that should be discussed.  That is, if $P(x)$ is the proposition "$x=2$" and $Q(x)$ is the proposition $(x-2)(x-3)=0$, the full-fledged logical statements whose true/false status is in question are
$$\forall x(P(x)\implies Q(x))\quad\text{and}\quad \forall x(Q(x)\implies P(x))$$
The first of these is true, the second is false.  The reason the second is false is simple:  $Q(3)$ is true, but $P(3)$ is false, so the implication $Q(x)\implies P(x)$ is not true for all $x$.  (Note, the implication is true for all $x$ other than $3$, mostly because the proposition $Q(x)$ is false for most values of $x$.)
It's perhaps worth adding why the statement $\forall x(P(x)\implies Q(x))$ is true.  The propositions $P(2)$ and $Q(2)$ are both true, so the implication $P(2)\implies Q(2)$ is true.  For all $x$ other than $2$, the proposition $P(x)$ is false, which makes any implication beginning "$P(x)\implies\ldots$" automatically true.  Hence the implication $P(x)\implies Q(x)$ is true for all $x$.
A: $$(2-2)(2-3)=0(-1)=0$$
In what universe is that false?
A: In the logic of statement, $P \Rightarrow Q$ is only wrong when $P$ is wrong, $Q$ is true. You can read here for more details https://en.wikipedia.org/wiki/Material_conditional. (mainly on the Truth table section)
In your question, it is a statement depend on variable $x$, that means the characteristic of true or wrong of $P(x) \Rightarrow Q(x)$ depend on the value of $x$.
Go on more specific, $x=2 \Rightarrow (x-2)(x-3)=0$, we consider $x \in \mathbb{R}$ (of course you can deal with the case $x \in \mathbb{C}$)
$\bullet  x=2$, we get $2=2 \Rightarrow (2-2)(2-3)=0$, or $P(x), Q(x)$ are true, so the statement is true
$\bullet x \neq 2$, we conclude that  $P(x)$ is always wrong, so by the Truth table, the statement is true
Conlusion: $P(x) \Rightarrow Q(x)$ is always true
So, apply the same method, you can answer why we can't use the argument $Q(x) \Rightarrow P(x)$, because when $x=3$, $Q(x)$ is wrong, $P(x)$ is true, so the statement $Q(x) \Rightarrow P(x)$ is wrong, of course it is true for $x \neq 3$ but if you want to use this, you must restrict to condition $x\neq 3$.
A little further: this question remind me of the reasonable transformations in solving an equation. That is  $P(x) =0 \Rightarrow Q(x)=0$ if and only if the solution set of $P(x)$ is a subset of the solution set of $Q(x)$, that means $P(x) \Rightarrow Q(x)$ is alwayse true if and only if the solution set of $P(x)$ is a subset of the solution set of $Q(x)$ (you can use the method above to prove that). When the solution set of $P(x)$ is exactly the same as the solution set of $Q(x)$, we can write $P(x)\Leftrightarrow Q(x)$. That make sense because when you solve $Q(x)$, there are some solution of $Q(x)$ that not the solution of $P(x)$ and you have to check again (this section is one of the most section that high school student often forget) 
A: Usually in logic, we talk about truth value for sentences, which are formulas without free variables. Here you have a free variable $x$. However, for any particular choice of $x$ the statement is true.
A: For the case $P(x)\Rightarrow Q(x)$:
If we choose $x=2$ then we got true on both sides of the arrow. If we choose $x\neq 2$ then P(x) is false and Q(x) can be false (if $x\neq 3$) or true (if $x=3$).
The implication $P(x)\Rightarrow Q(x)$ is therefore valid.
Following explanation will explain why I was wrong. This is for the case $P(x)\Leftarrow Q(x)$:
If we choose $x=2$ then we got true on both sides of the arrow.
If we choose $x=3$ then we got Q(x) is true and P(x) is false. Since this scenario is possible, it means the implication $P(x)\Leftarrow Q(x)$ is not valid, i.e. we have $P(x)\nLeftarrow Q(x)$.
A bit more intuitive and more self explaining similar example is when we consider $x=3 \Rightarrow x^2=9$ which is fully valid. What if we change the direction of the arrow in the implication? Changing direction of the arrow, gives us $x=3 \nLeftarrow x^2=9$ since $x=-3$ is true for $x^2=9$ but not for $x=3$.
