I have a somewhat very basic question. I have the following equations. Eq1 is the first raw form and then further simplying it leads to Eq2
My question is.
How come this happens since Eq2 came from Eq1?
Again, probably a very silly question.
On
It is not a silly a question.
Remember that you can only multiply an equation by a function that does not vanish in the domain of definition of the original equation if you are looking for equivalent transformations.
When you set $k = 0$ in the second equation, you obtain the solution $s = -3$.
But this is an extraneous root, because you had to assume that $s\neq -3$ in order to multiply both sides by the denominator given by $s^{2} + 6s + 9$. So, in both cases, setting $k = 0$ leads to an equation that has no solution.
Please let me know if you still need further assistance.
The issue that you raised occurs frequently in Algebra.
One starts with Constraint-1.
Then one determines that if Constraint-1 is satisfied, then Constraint-2 must also be satisfied.
Then, you determine all of the ways that Constraint-2 can be satisfied, and you discover that there is a way of satisfying Constraint-2 without satisfying Constraint-1.
Some people refer to this as Constraint-2 having extraneous roots.
My wording is simply that the navigation between Constraint-1 and (the derived) Constraint-2 is often a one way implication.
That is, Constraint-1 $~\implies~$ Constraint-2,
rather than Constraint-1 $~\iff~$ Constraint-2.
The classic example of this is
So, when starting with Constraint-1, and deriving Constraint-2, and then determining all of the solutions to Constraint-2, these solutions are each merely candidate solutions to Constraint-1. Each candidate solution must be manually checked against Constraint-1.
However, assuming that Constraint-2 has been derived from Constraint-1, you do know that no solution to Constraint-1 is possible unless it is one of the candidate solutions to Constraint-2.