# How can we find out the interval in an inequality?

Please go through the following link: Why is equating one of the bracks to zero in this equation correct?

Now, the expression given there is $(x+1)(x+3)$, I understand now why we take either of these two be zero and then find out the value. However, suppose I have the inequality: $(x + 2)(x - 3)\lt0$. In thise case, I'm expected to find out two values by evaluation $(x + 2)=0$ and $(x - 3)=0$ and then divide the real number line in 4 intervals and figure out for which intervals the inequality holds by figuring out whether the expression on the LHS is less than $0$ or greater than $0$.

However, what I don't get is, why are we allowed to use the evaluations $(x + 2)=0$ and $(x - 3)=0$? I understand that in the first case (in case of the equation), either of those values can be zero. But in this case, neither of values need to be zero in order to satisfy the inequality. Then how is equating to zero and finding out the value(s) of $x$ a valid operation?

• $0$ is very special in this regard. Knowing that $ab=0$ tells us that (at least) one of $a,b$ is $0$. This is an important property of the reals (or rationals, or integers).
– lulu
Apr 23, 2016 at 15:23
• But in this case, it's not an equation, it's an inequality. How can we know that either of those values is equal to zero? Apr 23, 2016 at 15:23
• Because, if one of the factors were $0$ the product would be $0$.
– lulu
Apr 23, 2016 at 15:24
• But the product is not zero, it's less than zero. Apr 23, 2016 at 15:24
• You've lost me. There are only two ways to have $ab<0$. Either $a<0 \;\&\;b>0$ or $a>0\;\&\;b<0$. For the second we need $x+2>0\;\&\;x-3<0$. The first turns out to be impossible.
– lulu
Apr 23, 2016 at 15:27

For a continuous function $f(x)$ on $[a,b]$, let $$f(a)\gt0$$ and $$f(b)\lt0$$ or vice-versa. Then, by the intermediate value theorem, there exists $c\in(a,b)$ such that $f(c)=0$.

So, when a continuous function changes sign, the graph must pass through the $x$-axis where its value is zero.

Suppose $f(p)=0$ and $f(q)=0$, $p\lt q$, such that $f(x)\ne0$ in $(p,q)$. Then, in the interval $(p,q)$, $f$ must either be positive throughout or negative throughout. If this were not true, then, as seen above, there would be some $p\lt r\lt q$ such that $f(r)=0$. So, the zeroes of $f$ demarcate the interval into regions where it is either only positive or only negative. This makes it easy to check the sign by simply finding the sign of the function at any convenient point in the interval.

• The question's tagged "precalculus" . Chances are the asker's knowledge of continuity, the IVT and etc. is either pretty basic or maybe even inexistent. Apr 23, 2016 at 15:30
• @Joanpemo You're right to make such an assumption. However, I was just brushing up my skills and this doubt came, so I asked. I have studied a little bit of Calculus. Apr 23, 2016 at 15:32
• @user331377 Good. Perhaps next time you can also add some few lines explaining your background. Apr 23, 2016 at 15:35

In high school we learn that $\;y=(x+1)(x+3)=x^2+4x+3\;$ is the equation of a (vertical) parabola opening upwards, and thus it is negative (meaning $\; (x+1)(x+3)<0\;$ which is precisely when its graph is below the $\;x-$ axis) exactly between its roots, meaning $\;y<0\iff -3<x<-1\;$ , and positive otherwise.

Using this very beautiful relation between algebra and geometry makes things much simpler, in my opinion, and also more intuitive.

For zero: $\;(x+1)(x+3)=0\;$ it is way simpler: we know that $\;ab=0\iff a=0\;$ or $\;b=0\;$ , that's why we do what you say we do to solve these equalities

• This is an inequality. Apr 23, 2016 at 15:25
• @user331377 That is exactly what I adresss in the first part of my answer. Read it all! Apr 23, 2016 at 15:26
• Got it, thanks a ton! Apr 23, 2016 at 15:28

By solving $(x+2)(x-3) = 0$, we determine where $(x+2)(x-3)$ changes sign; determining its sign in one of the intervals then determines its sign in all intervals, and hence the solution to $(x+2)(x-3) < 0$.

EDIT: to answer the question, solving the equality is a different problem, but useful for solving the corresponding inequality.

• So, the fact that we're using $(x + 2) = 0$ and $(x - 3) = 0$, doesn't mean it's true. It's just that we wanna find out where the sign changes to determine the intervals, correct? Apr 23, 2016 at 15:27
• @user331377 Indeed, we are just determining the points where the left hand side changes sign.
– user332714
Apr 23, 2016 at 15:28