[Common Math 1 Part 17] Interpreting and Solving Quadratic Inequalities

In the previous posts, we organized roots, the discriminant, graphs, and several forms of quadratic equations. Now we move one step further. Instead of asking only where an expression becomes 0, we ask where it is positive or negative.

Interpret quadratic inequalities through quadratic equations and graphs, and use roots and sign changes to find the solution set.

Let us fix the flow first.

• A quadratic equation asks where the expression becomes 0.
• A quadratic inequality asks on which intervals the expression is positive or negative.
• So after finding the roots, we must check how the sign changes around them.
• The graph helps us see why those intervals appear.

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1. What Is a Quadratic Inequality Asking?

For example,

$$x^2-5x+6=0$$

asks for the values of $x$ that make the expression equal to 0.

But

$$x^2-5x+6>0$$

asks for the range of $x$ where the expression is greater than 0.

So,

• an equation looks for points
• an inequality looks for intervals

That is the key difference.

This is why we usually begin by solving the related quadratic equation

$$x^2-5x+6=0$$

and then checking the sign in the intervals determined by its roots.

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2. Why Do We Find the Roots First?

The sign of a quadratic expression does not usually change at random. The important boundary points are where the expression becomes 0, that is, the roots.

For example,

$$x^2-5x+6=(x-2)(x-3),$$

so the roots are $x=2$ and $x=3$.

That naturally splits the number line into three intervals:

• $x<2$
• $2<x<3$
• $x>3$

Now we only need to determine the sign on each interval.

2-1. Why Is the Sign Constant Inside One Interval?

As long as we do not cross a root, the expression does not pass through 0. So within a single interval, a positive value cannot suddenly become negative.

From the graph point of view, a polynomial graph is continuous, so its sign cannot flip without crossing the $x$-axis.

In that sense, roots act as the boundary points that divide the sign behavior.

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3. When Factoring Works, a Sign Table Solves It Quickly

Consider

$$x^2-5x+6>0.$$

Factoring gives

$$(x-2)(x-3)>0.$$

Now we examine the sign of each factor.

3-1. Check the Sign by Interval

At first, it is perfectly fine to use one test value from each interval.

For example, we may choose

• $x=0$ in the region $x<2$
• $x=2.5$ in the region $2<x<3$
• $x=4$ in the region $x>3$

If we substitute $x=0$, then

$$(0-2)(0-3)=(-)(-)=(+).$$

Now the interval signs can be summarized as follows:

• if $x<2$, then $x-2<0$ and $x-3<0$, so the product is positive
• if $2<x<3$, then $x-2>0$ and $x-3<0$, so the product is negative
• if $x>3$, then both factors are positive, so the product is positive

Therefore the solution is

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

3-2. If Equality Is Included, the Endpoints Change Too

If the inequality is

$$x^2-5x+6\ge 0,$$

then we must also include the points where the expression equals 0.

That is because at the roots the expression is exactly 0, and $0\ge0$ is true.

So the solution becomes

$$x\le 2 \qquad \text{or} \qquad x\ge 3.$$

So:

• with $>$ or $<$, the roots are not included
• with $\ge$ or $\le$, the roots are included

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4. The Graph Shows Why the Outside Intervals Work

Now look at the same expression as a function:

$$y=x^2-5x+6.$$

This graph meets the $x$-axis at $x=2$ and $x=3$, and it opens upward.

So:

• for $x<2$ and $x>3$, the graph is above the $x$-axis
• for $2<x<3$, the graph is below the $x$-axis

That is why the solution of

$$x^2-5x+6>0$$

is the two outside intervals.

So a quadratic inequality can also be seen as

a problem of reading where the graph lies above or below the $x$-axis.

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5. If the Parabola Opens Downward, the Sign Pattern Can Reverse

Now consider

$$-x^2+5x-6>0.$$

Rewriting gives

$$-(x^2-5x+6)>0,$$

that is,

$$-(x-2)(x-3)>0.$$

The roots are the same as before, but the leading negative sign reverses the overall sign.

We can also see this algebraically: if we multiply both sides by $-1$, the inequality direction flips, so the problem becomes

$$x^2-5x+6<0.$$

So the solution is

$$2<x<3.$$

5-1. The Graph Makes This Simpler

The function

$$y=-x^2+5x-6$$

is a downward-opening parabola.

It meets the $x$-axis at $x=2$ and $x=3$, and between those two points it lies above the axis. That is why

$$-x^2+5x-6>0$$

keeps only the middle interval.

So whether the parabola opens upward or downward matters a lot when we read the solution set.

In the sign-chart explorer below, try switching the inequality sign and compare how the chosen interval changes even though the roots stay the same.

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6. What If There Is Only One Root or No Real Root?

The discriminant and the graph help us here as well.

6-1. The Repeated-Root Case

Consider

$$x^2-4x+4<0.$$

This becomes

$$(x-2)^2<0.$$

But a square is always at least 0, so it can never be less than 0. Therefore there is no solution.

On the other hand,

$$(x-2)^2\ge 0$$

is true for every real number $x$.

And if we ask

$$(x-2)^2\le 0,$$

then the only possible case is when the square is exactly 0, so the solution is

$$x=2.$$

In the repeated-root case, the graph only touches the $x$-axis once, so its position relative to the axis becomes especially clear.

6-2. The No-Real-Root Case

Now consider

$$x^2+1<0.$$

Since $x^2+1>0$ for every real number $x$, this inequality never holds. So there is no solution.

But

$$x^2+1>0$$

is true for all real numbers.

Here the leading coefficient is positive, so the parabola opens upward. Since there is no real root, the whole graph stays above the $x$-axis.

Now compare that with

$$-x^2-1>0.$$

This never holds, because $-x^2-1<0$ for every real number $x$. So even when there is no real root, the overall sign still depends on the sign of the leading coefficient.

So if there is no real root, the parabola never meets the $x$-axis, which means the whole graph stays on one side of the axis.

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7. The Basic Procedure for Solving Quadratic Inequalities

We can now summarize the process.

1. Find the roots of the related quadratic equation.
2. Use the roots to divide the number line into intervals.
3. Determine the sign of the expression on each interval.
4. Select the intervals that match the inequality.
5. Check whether the inequality includes equality, and decide whether to include the endpoints.

If you keep this procedure in mind, small changes in the appearance of the problem will not shake the method.

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8. Common Mistakes

8-1. Stopping with Only the Roots, as If It Were an Equation

A quadratic inequality is not asking for isolated roots. It is asking for intervals. So writing only $x=2,3$ is not enough.

8-2. Missing Whether Equality Is Included

$>$ and $\ge$ lead to different solution sets. We must check carefully whether the roots should be included.

8-3. Forgetting That a Leading Negative Reverses the Sign Pattern

For example,

$$-(x-2)(x-3)>0$$

must be read using the full sign, including the leading negative.

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9. Key Takeaways

• A quadratic equation finds points, but a quadratic inequality finds intervals.
• We usually solve a quadratic inequality by finding the roots first and then checking signs.
• In graph language, we are reading where the quadratic lies above or below the $x$-axis.
• Upward-opening and downward-opening parabolas can produce different solution-set shapes.
• $>$ and $<$ differ from $\ge$ and $\le$ because of endpoint inclusion.

In the next post, we will build on quadratic inequalities and move into more varied forms of inequalities.

One-line conclusion:

A quadratic inequality is not only about finding roots. It is about reading on which intervals the sign of the expression stays positive or negative, with the roots acting as boundaries.