Math Core

"When is the ball higher than 36 feet?" "For which prices is the profit positive?" Questions like these ask not where a quadratic equals a value, but where it is greater or less than it. The answer is an interval (or two), and the zeros of the quadratic mark where the intervals begin and end.

The idea: a parabola changes sign only at its zeros

Consider $f(x) = x^2 - x - 6 = (x - 3)(x + 2)$ . Its zeros are $-2$ and $3$ . The graph is a parabola opening up, so it dips below the x-axis between the zeros and stays above the x-axis outside them.

y = x² - x - 6 is negative between its zeros and positive outside them.Open in grapher →

So the solution of $x^2 - x - 6 < 0$ is $-2 < x < 3$ , and the solution of $x^2 - x - 6 > 0$ is $x < -2$ or $x > 3$ .

This works because a quadratic function is continuous: its graph has no breaks, so it can only change from positive to negative by passing through zero. The zeros cut the number line into pieces, and on each piece the sign never changes.

Solving a quadratic inequality

Move every term to one side, so the other side is $0$ .
Find the zeros of the quadratic (factor, or use the quadratic formula).
Use the direction of the parabola, or test one value in each interval, to see where the quadratic is positive and where it is negative.
Write the intervals that satisfy the inequality. Include the zeros for $\le$ or $\ge$ ; leave them out for $<$ or $>$ .

For a parabola that opens up ( $a > 0$ ) with two zeros, the pattern is always the same: negative between the zeros, positive outside. For $a < 0$ it's the reverse.

Worked example: Outside the zeros

Solve $x^2 + 2x \ge 8$ .

Move everything to one side: $x^2 + 2x - 8 \ge 0$ . Factor: $(x + 4)(x - 2) \ge 0$ , so the zeros are $-4$ and $2$ .

The parabola opens up, so it is positive outside the zeros. The inequality allows equality, so the zeros are included:

x \le -4 \quad\text{or}\quad x \ge 2.

Test points and negative leading coefficients

You don't have to picture the graph. Test one convenient number from each interval in the original inequality, and the sign you get holds for the whole interval.

Worked example: A test-point table

Solve $-2x^2 + 5x + 3 > 0$ .

Find the zeros of $-2x^2 + 5x + 3$ . Multiply by $-1$ to make factoring easier (this only changes the sign of the expression, not where it is zero): $2x^2 - 5x - 3 = (2x + 1)(x - 3)$ . The zeros are $-\dfrac{1}{2}$ and $3$ .

interval	test value	$-2x^2 + 5x + 3$	sign
$x < -\frac{1}{2}$	$x = -1$	$-2 - 5 + 3 = -4$	negative
$-\frac{1}{2} < x < 3$	$x = 0$	$3$	positive
$x > 3$	$x = 4$	$-32 + 20 + 3 = -9$	negative

The expression is positive only in the middle interval, so the solution is $-\dfrac{1}{2} < x < 3$ . That matches the picture: $a = -2 < 0$ means the parabola opens down, so it is above the axis between its zeros.

Common mistake

If you multiply or divide an inequality by a negative number, you must reverse the inequality sign. In the example above, rewriting $-2x^2 + 5x + 3 > 0$ as $2x^2 - 5x - 3 > 0$ (without flipping) would give exactly the wrong intervals. Either keep the original expression and use test points, or flip the sign: $2x^2 - 5x - 3 < 0$ .

When there are no real zeros

If the discriminant is negative, the parabola never touches the x-axis, so the quadratic has the same sign for every $x$ . Check its sign at any one point to see which.

For example, $x^2 + 2x + 5$ has $D = 4 - 20 = -16 < 0$ , and at $x = 0$ it equals $5 > 0$ . So $x^2 + 2x + 5$ is always positive:

$x^2 + 2x + 5 > 0$ is true for all real numbers.
$x^2 + 2x + 5 < 0$ has no solution.

A perfect square behaves similarly. $(x - 3)^2 \ge 0$ for all $x$ , so $(x - 3)^2 < 0$ has no solution, and $(x - 3)^2 \le 0$ has only the solution $x = 3$ .

An application

Worked example: Height above a target

A ball's height in feet after $t$ seconds is $h(t) = -16t^2 + 48t + 4$ . During what time interval is the ball higher than $36$ feet?

Solve $-16t^2 + 48t + 4 > 36$ :

\begin{aligned} -16t^2 + 48t - 32 &> 0 \\ t^2 - 3t + 2 &< 0 \qquad \text{(divide by } -16 \text{ and flip)} \\ (t - 1)(t - 2) &< 0 \end{aligned}

The parabola $y = t^2 - 3t + 2$ opens up, so it is negative between its zeros. The ball is above $36$ feet for $1 < t < 2$ , a span of $1$ second.

Inequalities in two variables

An inequality such as $y > x^2 - 4$ describes a region of the plane. Graph the boundary parabola $y = x^2 - 4$ , dashed for $<$ or $>$ and solid for $\le$ or $\ge$ , then shade the side that satisfies the inequality. For $y > \ldots$ that's the region above (inside) the parabola.

y > x² - 4: the dashed boundary is not included, and the shading lies above the parabola. (0, 0) is a solution; (3, 0) is not.Open in grapher →

To check which side to shade, test a point not on the boundary. For $(0, 0)$ : is $0 > 0^2 - 4 = -4$ ? Yes, so the region containing the origin is shaded. For $(3, 0)$ : is $0 > 9 - 4 = 5$ ? No.

Tip

A quick sanity check on any one-variable answer: pick a number your answer says is a solution and one it says is not, and substitute both into the original inequality. One should make it true and the other false.