[Common Math 1 Part 18] Various Inequalities

In the previous post, we organized quadratic inequalities through roots and graphs. Now we widen that idea a little and study several inequalities that look different on the surface, but still reduce to the same core task: reading signs.

Interpret product inequalities and rational inequalities through sign charts, and understand especially how to treat points where the denominator becomes zero.

Let us fix the flow first.

• Various inequalities still ask whether an expression is positive or negative.
• So the important points are where the expression becomes 0 and where it is undefined.
• Those points divide the number line into intervals.
• In rational inequalities especially, a point that makes the denominator 0 can never be part of the solution set.

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1. What Do "Various Inequalities" Really Ask?

For example,

$$(x-1)(x+2)>0$$

and

$$\frac{(x-1)(x+2)}{x-3}>0$$

do not look the same.

But both problems are asking the same question:

On which intervals is this expression positive?

So the key is not to start calculating blindly, but to find first

• where the value becomes 0
• where the expression is undefined

Those are the boundary points where the sign behavior can change.

Here, an expression is undefined mainly in rational expressions: if the denominator becomes 0, the value cannot be computed.

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2. In a Product Inequality, Read the Signs of the Factors

First consider

$$(x-1)(x+2)>0.$$

This expression becomes 0 at

$$x=1, \qquad x=-2.$$

So the number line is divided into three intervals:

• $x<-2$
• $-2<x<1$
• $x>1$

2-1. Check the Sign by Interval

We can use one test value in each interval.

• if $x=-3$, then $(-4)(-1)>0$
• if $x=0$, then $(-1)(2)<0$
• if $x=2$, then $(1)(4)>0$

Therefore the solution is

$$x<-2 \qquad \text{or} \qquad x>1.$$

2-2. Why Does This Work?

The sign of a product is determined by the signs of its factors.

• positive × positive = positive
• negative × negative = positive
• positive × negative = negative

So a product inequality is really a problem of reading the sign combination of the factors.

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3. If Equality Is Included, the Zeros Matter Too

Now consider

$$(x-1)(x+2)\le 0.$$

From the sign analysis above,

• the expression is negative on $-2<x<1$
• and it is 0 at $x=-2$ and $x=1$

So the solution is

$$-2\le x\le 1.$$

The important distinction is this:

• with $<$ or $>$, points where the value is 0 are not included
• with $\le$ or $\ge$, points where the value is 0 are included

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4. A Rational Inequality Requires One More Check

Now look at

$$\frac{(x-1)(x+2)}{x-3}>0.$$

This looks similar to a product inequality, but there is one more thing to watch:

The sign-reading method itself is still the same as before. The new issue is simply that we now have a denominator.

the point where the denominator becomes 0.

Here,

• the numerator becomes 0 at $x=1$ and $x=-2$
• the denominator becomes 0 at $x=3$

So the number line is divided into four intervals:

• $x<-2$
• $-2<x<1$
• $1<x<3$
• $x>3$

4-1. Why Must We Mark $x=3$ Separately?

At $x=3$, the denominator is 0, so the expression is undefined.

That means

$$\frac{(x-1)(x+2)}{x-3}$$

has no value at $x=3$.

So $x=3$ is both a boundary point for sign changes and a point that can never belong to the solution set.

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5. A Rational Inequality Is Also Solved by a Sign Chart

Now let us check the sign interval by interval.

5-1. Use Test Values Again

Let us write one interval out in full first.

At $x=0$,

$$\frac{(0-1)(0+2)}{0-3}=\frac{(-)(+)}{(-)}=\frac{(-)}{(-)}=(+).$$

Now the other intervals can be checked in the same way:

• if $x=-3$, then $\frac{(-)(-)}{-}<0$
• if $x=0$, then $\frac{(-)(+)}{-}>0$
• if $x=2$, then $\frac{(+)(+)}{-}<0$
• if $x=4$, then $\frac{(+)(+)}{+}>0$

Therefore,

$$\frac{(x-1)(x+2)}{x-3}>0$$

has solution

$$-2<x<1 \qquad \text{or} \qquad x>3.$$

Here, $x=-2$ and $x=1$ make the expression equal to 0, so they are not included because the inequality is strict. And $x=3$ is not included because the expression is undefined there.

5-2. The Sign Chart Makes the Structure Visible

We can read the same problem like this:

$$\begin{array}{c|cccc} x & x<-2 & -2<x<1 & 1<x<3 & x>3 \\ \hline x+2 & - & + & + & + \\ x-1 & - & - & + & + \\ x-3 & - & - & - & + \\ \hline \text{overall sign} & - & + & - & + \end{array}$$

So a rational inequality is solved in almost the same way as a product inequality. The extra step is that points making the denominator 0 must be treated separately.

In the sign-chart explorer below, try switching the inequality sign. Pay special attention to what happens near $x=3$, and notice why the denominator-zero point can never belong to the solution set.

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6. The Most Important Exception in Rational Inequalities: The Denominator Cannot Be 0

Consider

$$\frac{x-1}{x-3}\ge 0.$$

At $x=1$, the value is 0, so that point may be included. But at $x=3$, the denominator is 0, so that point cannot be included.

So we must distinguish carefully between

• a point where the numerator is 0
• a point where the denominator is 0

For example, in $\frac{x-1}{x-3}\ge0$, the point $x=1$ may be included, but $x=3$ is excluded even though the inequality is non-strict.

6-1. Why Are They Treated Differently?

If the numerator is 0, then the whole expression may become 0. So for $\ge 0$ or $\le 0$, such a point may belong to the solution set.

But if the denominator is 0, the expression is not defined at all. An undefined value can never belong to the solution set.

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7. The Common Procedure for Various Inequalities

We can now summarize the method.

1. Rewrite the expression into a form where the sign is easy to read.
2. In a product inequality, find the zeros; in a rational inequality, find both numerator zeros and denominator zeros.
3. Use those points to divide the number line into intervals.
4. Determine the overall sign on each interval.
5. Select the intervals that satisfy the inequality.
6. Decide endpoint inclusion by checking whether the point gives 0 or makes the expression undefined.

If you remember this procedure, even a more complicated-looking inequality becomes much more manageable. The product example and the rational example above both follow exactly this same order.

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8. Common Mistakes

8-1. Including a Point Where the Denominator Is 0

This is the most common and most important mistake in rational inequalities.

Even if the inequality is $\ge$ or $\le$, a point that makes the denominator 0 can never be included.

8-2. Testing Only One Interval and Guessing the Rest

The sign may change from interval to interval, so each interval must be checked separately.

8-3. Treating Numerator Zeros and Denominator Zeros the Same Way

A numerator zero may be included, depending on the inequality. A denominator zero is never included.

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9. Key Takeaways

• Various inequalities are still problems of reading the sign of an expression.
• Product inequalities are solved by reading the signs of the factors.
• Rational inequalities use the same idea, but denominator-zero points must be handled separately.
• The numerator may become 0, but the denominator may not.
• The solution set depends on both sign behavior and whether each boundary point is defined.

In the next post, we will briefly organize the whole equations-and-inequalities block before moving into the next major unit of Common Math 1.

One-line conclusion:

Even when inequalities look different on the surface, they are still problems of reading signs, using zeros and undefined points as the boundaries.