# Should $x=-2$ be included as an answer for $\frac{x^2+8x+12}{x^2+5x+6}>0$?

$$\frac{x^2+8x+12}{x^2+5x+6}>0$$ First of all while solving inequalities I need to check domain so in this case $$x^2+5x+6\neq0$$ $$x\neq-2,\ x\neq-3$$ Later on $$\frac{(x+6)(x+2)}{(x+3)(x+2)}>0$$ Then get critical values draw number line and get $$x\in(-\infty;-6)\cup(-3;-2)\cup(-2;+\infty)$$ However according to wolframalpha $x=-2$ is included as an answer.

So am I wrong or wolframalpha is wrong?

Also I checked $\frac{x}{x}=1$ and wolframalpha also includes $x=0$ but once again I think it's incorrect?

• Well, there is a removable singularity at $x=-2$. – copper.hat Jan 2 '13 at 1:47
• Try solving $x/x > 0$ with Wolfram. However, $1/x > 0$ produces the correct answer. – copper.hat Jan 2 '13 at 1:51
• I am inclined to agree with Wolfram Alpha. Your expression "really wants" to be $\frac{x+6}{x+3}$, and perhaps assumed its more convoluted form by accident. With trigonometric identities, one tends to take this poin of view. For example, there is a standard identity that expresses $\sum_{k=1}^n \cos k\theta$ as a fraction with $\sin(\theta/2)$ in the denominator. The identity is technically incorrect when $\sin(\theta/2)=0$, but it is standard not to worry about it. – André Nicolas Jan 2 '13 at 1:58
• @copper.hat ok so it's that way probably because wolfram simplifies expression at the beginning, but why it does so, if it's not always correct? why it doesnt check for a domain first? – Templar Jan 2 '13 at 2:01
• I am a little surprised that Wolfram doesn't indicate that it has removed the removable singularities first. – copper.hat Jan 2 '13 at 2:05

As it seems you are simply solving an inequality, and entered it as such, you are correct: the expression is not defined at $-2$.

Why Wolfram Alpha omitted $\,-3\,$ from the solutions, but included $\,-2\,$, seems inconsistent to me, as you note! It seems that if Wolfram is going to at least be consistent...either both values should be omitted, or both included.

Wolfram alpha probably simplified the expression before it calculated the solutions. You are right. It is worth mentioning, however, that there is a hole at $x=-2$, and the limit as $x\rightarrow -2$ satisfies the inequality.

• so is it a bug? can't believe noone noticed it for such long time – Templar Jan 2 '13 at 1:49
• Yeah, I guess it must be. Perhaps alpha is programmed to include limit points in its solution sets automatically - your $x/x$ example suggests the latter. – Alexander Gruber Jan 2 '13 at 1:54
• @AlexanderGruber This is probably not the explanation. Try solving $1/x^2>0$. – mrf Jan 2 '13 at 8:15
• @Templar: can't believe noone noticed it for such long time... An awful lot of people did notice it, and many other features of W|A, for a very long time. In fact these are quite well known. The trouble begins when people consider W|A as something it is not. – Did Jan 2 '13 at 10:03

The function as written is certainly not defined at $x= -2,$ but it does have a removable singularity there. The only way to extend the definition (note the word extend) of the function so that it becomes continuous at $x = -2$ is to define the value at $-2$ to be $4$. Then the new function is identically equal to $\frac{x+6}{x+3}$ except at $x = -3,$ where it is not defined (and worse, the singularity at $x = -3$ is not removable).

No, your expression is undefined at $x = -2$. Are you taking a limit from the left or right???

• why wolframalpha gives -2 as an answer then? – Templar Jan 2 '13 at 1:43
• It's kinda hairsplittery. I think that the expression is undefined there. Wolfram is eliding the removable singularity. I think it's a matter of taste, but I say $x \not= -2$. – ncmathsadist Jan 2 '13 at 1:45
• Wolfram is wrong, but justifiably so. – copper.hat Jan 2 '13 at 1:48
• Wolfram is not perfectt. – copper.hat Jan 2 '13 at 1:53