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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

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My question is.

  • If I substitute k=0 in Eq1, I get 0 = -1
  • If I substiute k=0 in Eq2, I am able to solve the quadratic equation in s.

How come this happens since Eq2 came from Eq1?

Again, probably a very silly question.

AlfroJang80
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    The original equation is what counts: $;k;$ cannot be zero as then the left side vanishes, what would render the equation as false. Observe also that it must be $$s^2+6x+9=(s+3)^2\neq0\iff s\neq -3$$ otherwise you divide by zero... – DonAntonio Mar 24 '22 at 18:43

2 Answers2

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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.

Átila Correia
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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

  • Constraint-1 : $x = 2.$
  • Constraint-2 : $x^2 = 4.$

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.

user2661923
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  • Understood. In future, when do I know that I need to recheck my Eq2 solution with Eq1 to see if it is valid? In other words what operation did I perform between Eq1 and Eq2 such that it generated a new constraint ? – AlfroJang80 Mar 24 '22 at 19:13
  • It takes years of advanced study/experience to distinguish, at-a-glance, which derivations are one-way versus two-way implications. The easiest course is to always assume that each Constraint-2 solution must always be checked against Constraint-1. Note that this issue also arises when constructing an $\epsilon,\delta$ proof of continuity. One starts with Constraint-1, and then derives that a specific Constraint-2 association between $\epsilon$ and $\delta$ seems pertinent. However, to be valid, you then have to manually verify that the Constraint-2 association satisfies Constraint-1. – user2661923 Mar 24 '22 at 19:25