Functions and Quadratic Inequalities

Try this

Switch between $> 0$ and $< 0$ . Where does the shading move? The two solutions together cover almost the whole $x$ -axis. Which two points are missing?
Switch to $\geq 0$ . What happens to the little circles at the endpoints? Why?
Drag $a$ below zero. It is still “ $> 0$ ”, but the solution moves from “the two outside pieces” to “the middle”. What happened to the graph?
Make $a$ positive again and drag $c$ upward until the graph leaves the $x$ -axis. What is the solution of $> 0$ now? And of $< 0$ ?

Solving $a x^{2} + b x + c > 0$ just means asking: which part of the graph is above the $x$ -axis? The $x$ values directly below that part are the solution.

Predict first

What is the solution of the inequality $x^{2} > 4$ ?

x > 2

x > \pm 2

x < - 2

x > 2

- 2 < x < 2

Choose an answer first; then you can see the result.

Why: a quadratic inequality in three steps

Example: solve x² − x − 6 < 0

Step 1 / 4

Step 1: see it as a function
Let $y = x^{2} - x - 6$ . The inequality asks: which $x$ make $y$ negative? For example, at $x = 1$ we get $y = - 6$ , so that one counts.

$y = x^{2} - x - 6$
Step 2: solve the equation to find the boundary points
Step 3: see which side the graph is on
Try a different sign

Many people memorize “greater than: the outside; less than: the middle”. That is only true when $a > 0$ and there are two different real roots. Draw the picture and you will know when it applies.

When a is negative

One problem, several solutions

Solve the inequality $- x^{2} + 2 x + 3 > 0$ .

Read it from the picture

$y = - x^{2} + 2 x + 3$ opens downward and meets the $x$ -axis at $- 1$ and $3$ . For a downward parabola, the part above the $x$ -axis is the arch between the roots: $- 1 < x < 3$ .

CostNo rewriting, so the sign cannot go wrong; you must draw the opening the right way.

Multiply by −1 first

Multiply both sides by $- 1$ and flip the sign: $x^{2} - 2 x - 3 < 0$ . It opens upward, and “less than” means between the roots: $- 1 < x < 3$ .

CostTurns it into the familiar upward case; the sign must flip, and that is the easiest step to get wrong.

The two pictures are upside down relative to each other, yet the thick line on the $x$ -axis is exactly the same. That is what “multiply by $- 1$ and flip the sign” looks like.

Holding for every x

If the whole parabola floats above the $x$ -axis, then $a x^{2} + b x + c > 0$ holds for every $x$ . That needs two things: it opens upward ( $a > 0$ ), and it never reaches the $x$ -axis ( $Δ < 0$ ).

x^{2} + 2 x + 3 > 0

holds for all real numbers. Drag

c

down: at which notch does it stop holding for all x?

Example: $x^{2} - 2 x + m > 0$ holds for every real $x$ ; find the range of $m$ . It already opens upward, so we only need $Δ = 4 - 4 m < 0$ , that is, $m > 1$ . Another way to see it: $x^{2} - 2 x + m = (x - 1)^{2} + (m - 1)$ has minimum $m - 1$ . If even the minimum is greater than $0$ , everything else is bigger still.

Remove one condition

Break it

Counterexample lab: remove one condition

Conclusion: The solution of $a x^{2} + b x + c > 0$ lies outside the roots (“greater than: the outside”)

When $a > 0$ the parabola opens upward, and the part above the $x$ -axis really is the two pieces outside the roots.

Practice

Exercise 1Basic

How many integer solutions does the inequality $x^{2} - 5 x + 4 < 0$ have?

Exercise 2Basic

Among the $x$ satisfying $x^{2} - 3 x - 10 \leq 0$ , what is the largest?

Exercise 3Advanced

How many integer solutions does the inequality $- x^{2} + 4 x + 5 > 0$ have?

Exercise 4Advanced

The solution of the inequality $x^{2} + b x + c < 0$ is exactly $- 1 < x < 4$ . Write $x^{2} + b x + c$ .

Exercise 5Challenge

$x^{2} - 2 k x + k + 2 > 0$ holds for every real $x$ . How many integers $k$ are there with this property?

Try this

Why: a quadratic inequality in three steps

Example: solve x² − x − 6 < 0

When a is negative

Read it from the picture

Multiply by −1 first

Holding for every x

Remove one condition

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Multiplying by −1 without flipping the inequality sign

Taking the square root of x² > 4 as if it were an equation

“Holds for all x”, missing the case where the leading coefficient is 0