Algebra 2

Quadratic inequalities

You can find where a quadratic equals zero. How do you find everywhere it is bigger than zero, or smaller?

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Roots, zeros and x-interceptsAlgebra 1 Compound inequalities: and, orAlgebra 1

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Polynomial and rational inequalities with sign chartsPrecalculus

Where you are 12 in Quadratics

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Roots, zeros and x-intercepts

Quadratic inequalities

Quadratic expressions and parabolas4 Solving quadratics by factoring Solving quadratics with square roots ▾ Show all 12 (7 more)

When you reach for it

A linear inequality such as 2x - 6 > 0 has one boundary, x = 3, and everything on one side of it works. A quadratic inequality, such as

x^2 - 2x - 3 > 0

is different. Try a few numbers. x = 5 works, since 25 - 10 - 3 = 12. x = 0 does not, since it gives -3. And x = -2 works again, since 4 + 4 - 3 = 5. The numbers that work are not all on one side of anything. They come in stretches, and the job is to find where each stretch starts and stops.

You reach for this move whenever the question is “for which x is this quadratic positive, or negative?” rather than “for which x is it zero?”. That question turns up when a ball must be above a certain height, when an area must be at least some amount, when a square root must have something non-negative underneath it, and when an equation must have two real solutions. The answer is almost never a single number. It is a set of numbers, written in interval notation.

The procedure

The whole method rests on one fact about the graph of y = ax^2 + bx + c: it is a single unbroken curve. To get from above the x-axis to below it, the curve has to cross the axis, and it crosses only at its roots. So between two neighboring roots, and beyond the outermost ones, the quadratic keeps the same sign the whole way along. A quadratic has at most two real roots, so the number line breaks into at most three stretches, and each stretch is entirely in the answer or entirely out of it.

The graph of y = x^2 - 2x - 3 shows this at a glance.

Figure 1Where the graph is above the axis, x^2 - 2x - 3 > 0. Those stretches of the x-axis are the solution.

Solving x^2 - 2x - 3 > 0 means finding the x where the graph is above the axis. You do not need to draw the graph to do it. Here is the same work on the number line alone.

Get zero on one side. The comparison has to be with 0, because the sign of a quantity is what the roots tell you about. Here it already is.

Find the roots of the matching equation. Solve x^2 - 2x - 3 = 0 by whatever method suits, here factoring:

(x + 1)(x - 3) = 0

so x = -1 or x = 3. These are the only places the sign can change. They are called the critical points.

Mark them on a number line. They split it into three stretches: x < -1, -1 < x < 3 and x > 3.

Test one number from each stretch. Because the sign cannot change inside a stretch, one number speaks for all of it. Write f(x) = x^2 - 2x - 3.

\begin{aligned} f(-2) &= 4 + 4 - 3 = 5 &&> 0 \\ f(0) &= -3 &&< 0 \\ f(4) &= 16 - 8 - 3 = 5 &&> 0 \end{aligned}

Figure 2The same answer without a graph. One test point decides each stretch.

Read off the stretches that work, and decide the ends. The inequality asks for > 0, so the two outer stretches are in and the middle one is out. The critical points themselves make the quadratic exactly 0, which is not > 0, so they are left out. That is what the open circles say. The answer is a compound inequality with or:

x < -1 \quad\text{or}\quad x > 3

In interval notation, (-\infty, -1) \cup (3, \infty).

Two things speed this up once you trust it. First, the factors tell you the signs directly. In (x + 1)(x - 3), both factors are negative when x < -1, so the product is positive; one is negative and one positive between the roots, so the product is negative; both are positive when x > 3. That is the sign rule for products doing the test for you. Second, the shape tells you the pattern. When a > 0 the parabola opens upward, so it is below the axis between its roots and above it outside them. When a < 0 it is the other way round. A test point is still the safest check, because it cannot be misremembered.

Worked examples

Example 1

Between the roots, ends included

Solve x^2 + 2x - 15 \le 0. Give the answer in interval notation.

Solution

Zero is already on the right. Factor the matching equation:

(x + 5)(x - 3) = 0

so x = -5 or x = 3. Test one number in each of the three stretches:

\begin{aligned} x = -6&: \quad 36 - 12 - 15 = 9 &&> 0 \\ x = 0&: \quad -15 &&< 0 \\ x = 4&: \quad 16 + 8 - 15 = 9 &&> 0 \end{aligned}

The inequality wants the quadratic negative or zero, so the middle stretch is in, and so are both critical points, where the quadratic equals 0. The answer is -5 \le x \le 3, or [-5, 3].

The square brackets are the “or equal to” part. With < instead of \le the answer would be (-5, 3).

Example 2

Moving terms first, and an answer in two pieces

Solve x^2 \ge 4x + 12.

Solution

The right side is not zero, and the roots only tell you where something equals zero. Subtract 4x + 12 from both sides, which keeps the inequality true because subtracting the same quantity from both sides never changes which side is larger:

x^2 - 4x - 12 \ge 0

Factor: (x - 6)(x + 2) = 0, so the critical points are x = -2 and x = 6.

Test: x = -3 gives 9 + 12 - 12 = 9 > 0; x = 0 gives -12 < 0; x = 7 gives 49 - 28 - 12 = 9 > 0. The outer stretches are in, and the critical points are in too, because of the \ge:

x \le -2 \quad\text{or}\quad x \ge 6

In interval notation, (-\infty, -2] \cup [6, \infty). The round bracket stays at \infty because infinity is not a number that can be included.

Check. x = 6: 36 \ge 24 + 12 = 36, true, with equality. x = 10: 100 \ge 52, true.

Example 3

A negative leading coefficient

Solve 3 + 5x - 2x^2 \ge 0.

Solution

It is easier to factor with a positive x^2 term, so multiply both sides by -1. Multiplying an inequality by a negative number reverses its direction, because it swaps which side is larger:

2x^2 - 5x - 3 \le 0

Factor: (2x + 1)(x - 3) = 0, so the critical points are \displaystyle x = -\frac{1}{2} and x = 3.

Now a = 2 > 0, so this parabola opens upward and is below the axis between its roots. Check it with a test point anyway: x = 0 gives -3 \le 0, true. Outside, x = 4 gives 32 - 20 - 3 = 9, which is not \le 0. So the answer is

-\tfrac{1}{2} \le x \le 3

In interval notation, \displaystyle \left[-\frac{1}{2}, 3\right].

You can also leave the original alone and think about its graph. y = 3 + 5x - 2x^2 opens downward, so it is above the axis between its roots, which is exactly what \ge 0 asks for. Both ways give the same stretch, as they must, since multiplying by -1 and flipping the sign describe the same set of numbers.

Figure 3Example 3. A parabola that opens downward is above the axis between its roots.

Example 4

Roots that are not whole numbers

Solve x^2 - 4x + 1 > 0.

Solution

This does not factor over the integers, so find the critical points with the quadratic formula, with a = 1, b = -4, c = 1:

\begin{aligned} x &= \frac{4 \pm \sqrt{16 - 4}}{2} \\ &= \frac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3} \end{aligned}

To choose test points you need to know roughly where these are. Since \displaystyle \sqrt{3} \approx 1.73, the roots are about 0.27 and 3.73. So 0, 2 and 4 each sit in a different stretch:

\begin{aligned} x = 0&: \quad 1 &&> 0 \\ x = 2&: \quad 4 - 8 + 1 = -3 &&< 0 \\ x = 4&: \quad 16 - 16 + 1 = 1 &&> 0 \end{aligned}

The outer stretches are in, and the ends are out because the inequality is strict. Keep the exact values in the answer; the decimals were only for placing test points:

x < 2 - \sqrt{3} \quad\text{or}\quad x > 2 + \sqrt{3}

In interval notation, \displaystyle (-\infty, 2 - \sqrt{3}) \cup (2 + \sqrt{3}, \infty).

Example 5

No real roots at all

Solve (a) x^2 + 2x + 5 > 0 and (b) x^2 + 2x + 5 < 0.

Solution

The matching equation x^2 + 2x + 5 = 0 has discriminant b^2 - 4ac = 4 - 20 = -16, which is negative, so there are no real roots. With no critical points there is nowhere for the sign to change, and the whole number line is one stretch. One test point settles everything: x = 0 gives 5 > 0.

Completing the square shows why, with no test point needed:

x^2 + 2x + 5 = (x + 1)^2 + 4

A square is never negative, so this is always at least 4.

(a) Every real number works: (-\infty, \infty).

(b) No real number works. The inequality has no solution.

Example 6

One root, touching the axis

Solve (a) x^2 - 6x + 9 \le 0 and (b) x^2 - 6x + 9 > 0.

Solution

The quadratic is a perfect square:

x^2 - 6x + 9 = (x - 3)^2

A square is zero when its inside is zero and positive otherwise. So this quadratic is 0 at x = 3 and positive at every other x. Its graph touches the axis at 3 and turns back up.

(a) (x - 3)^2 \le 0 asks for negative or zero. Nothing is negative, and zero happens only at x = 3. The solution is the single number x = 3.

(b) (x - 3)^2 > 0 holds everywhere except at x = 3:

(-\infty, 3) \cup (3, \infty)

Here the one critical point splits the line into two stretches with the same sign. The sign changes at a root only when the curve crosses the axis, not when it just touches it.

Example 7

Using it: when does an equation have two solutions?

For which values of k does x^2 + kx + k + 3 = 0 have two different real solutions?

Solution

A quadratic equation has two different real solutions exactly when its discriminant is positive. Here a = 1, b = k and c = k + 3, so the condition is

\begin{aligned} k^2 - 4(k + 3) &> 0 \\ k^2 - 4k - 12 &> 0 \end{aligned}

That is a quadratic inequality, with k as its variable. Its critical points come from k^2 - 4k - 12 = (k + 2)(k - 6) = 0, so k = -2 or k = 6. Test the middle stretch: k = 0 gives -12, negative, so it is out, and the two outer stretches are in. The ends are out because the inequality is strict.

k < -2 \quad\text{or}\quad k > 6

Check. With k = 7 the equation is x^2 + 7x + 10 = (x + 2)(x + 5) = 0, two solutions. With k = 6 it is x^2 + 6x + 9 = (x + 3)^2 = 0, only one, which is why 6 itself is left out. With k = 0 it is x^2 + 3 = 0, which has none.

Boundaries

The comparison must be with zero. Roots say where a quadratic equals 0, so they only split ax^2 + bx + c > 0, not ax^2 + bx + c > 7. Move everything to one side first, as in Example 2. In particular, x^2 < 5x cannot be divided through by x: x might be negative, which would flip the sign, or zero, which cannot be divided by at all. Write x^2 - 5x < 0 instead.

Multiplying by a negative flips the inequality. It is allowed, and often convenient, but the direction has to change with it, as in Example 3.

The sign changes only where the curve crosses. Two different real roots give three stretches with alternating signs. A repeated root gives two stretches with the same sign. No real roots give one stretch, so the answer is every real number or none.

Figure 4With a > 0, the three pictures a quadratic can make. Only the first ever goes negative.

The ends follow the inequality sign. With < or > the critical points are left out; with \le or \ge they are in. A critical point always makes the quadratic exactly 0, so this is the only thing that decides it.

The method needs an unbroken curve. Every polynomial has one, which is why the same test-point method works for a cubic. It fails for an expression such as \displaystyle \frac{1}{x - 2}, which changes sign at x = 2 without ever being zero there; for a fraction, the places where the denominator is zero have to be added to the critical points.

Where this goes next

The same idea, sign changes only at zeros, is how you find where any function is positive or negative, and, applied to a derivative in calculus, it becomes the sign chart that tells you where a function is rising and where it is falling.

Quadratic inequalities hide inside other questions. Finding the domain of \displaystyle \sqrt{8 + 2x - x^2} is one, since what is under a square root must be at least 0. Asking when an equation has two real roots is another, as in Example 7. And in projectile problems, “when is the ball higher than 40 meters?” is a quadratic inequality in time.

It also connects back to absolute value. x^2 < 9 has the solution -3 < x < 3, which is exactly the solution of the absolute value inequality |x| < 3: a square is less than 9 precisely when the number is closer to 0 than 3 is.

Exercises

Solve by factoring

Solve each inequality. Give the answer in interval notation.

x^2 - 5x + 4 < 0
x^2 + 3x - 10 \ge 0
x^2 - 9 > 0
x^2 + 6x \le 0
x^2 < 2x + 15
x^2 + 7x + 12 > 0
-x^2 + x + 20 \ge 0
2x^2 + x - 6 > 0
6 - x - x^2 < 0
3x^2 \le 10x + 8

Roots that are not whole numbers

x^2 - 2x - 1 < 0
x^2 + 4x + 1 \ge 0
x^2 > 6x - 4

No real roots, or only one

x^2 + x + 1 > 0
x^2 - 4x + 7 \le 0
x^2 + 10x + 25 > 0
4x^2 - 12x + 9 \le 0

Using it

A ball is thrown straight up, and its height after t seconds is h = 30t - 5t^2 meters. For which times is the ball higher than 40 meters?
For which values of k does x^2 + kx + 4 = 0 have two different real solutions?
For which x is \displaystyle \sqrt{8 + 2x - x^2} a real number?
A rectangle has a perimeter of 40 cm. If one side is x cm, for which x is the area at least 96 square centimeters?
Write a quadratic inequality, with x^2 as its first term, whose solution is exactly (-\infty, -1) \cup (4, \infty).

Reasoning

Explain why, once you know the critical points, testing a single number in each stretch is enough to decide the whole stretch.
Explain why no quadratic inequality can have the solution (0, 1) \cup (2, 3).
Show that x^2 < 9 and |x| < 3 have the same solutions.

Answers

(1, 4). (x - 1)(x - 4) < 0; x = 2 gives -2, so the middle stretch.
(-\infty, -5] \cup [2, \infty). (x + 5)(x - 2) \ge 0; x = 0 gives -10, so the outer stretches, ends included.
(-\infty, -3) \cup (3, \infty). (x + 3)(x - 3) > 0; x = 0 gives -9, so the outer stretches.
[-6, 0]. x(x + 6) \le 0; x = -1 gives -5, so the middle stretch, ends included.
(-3, 5). x^2 - 2x - 15 < 0, so (x - 5)(x + 3) < 0; x = 0 gives -15.
(-\infty, -4) \cup (-3, \infty). (x + 3)(x + 4) > 0; x = -3.5 gives -0.25, so the middle stretch is out.
[-4, 5]. Multiply by -1 and flip: x^2 - x - 20 \le 0, so (x - 5)(x + 4) \le 0; x = 0 gives -20.
\displaystyle (-\infty, -2) \cup \left(\frac{3}{2}, \infty\right). (2x - 3)(x + 2) > 0; x = 0 gives -6, so the middle stretch is out.
(-\infty, -3) \cup (2, \infty). Multiply by -1 and flip: x^2 + x - 6 > 0, so (x + 3)(x - 2) > 0; x = 0 gives -6.
\displaystyle \left[-\frac{2}{3}, 4\right]. 3x^2 - 10x - 8 \le 0, so (3x + 2)(x - 4) \le 0; x = 0 gives -8.
\displaystyle (1 - \sqrt{2}, 1 + \sqrt{2}). The quadratic formula gives \displaystyle x = 1 \pm \sqrt{2}, about -0.41 and 2.41; x = 0 gives -1, so the middle stretch.
\displaystyle (-\infty, -2 - \sqrt{3}] \cup [-2 + \sqrt{3}, \infty). The roots are \displaystyle -2 \pm \sqrt{3}, about -3.73 and -0.27; x = -2 gives -3, so the middle stretch is out.
\displaystyle (-\infty, 3 - \sqrt{5}) \cup (3 + \sqrt{5}, \infty). x^2 - 6x + 4 > 0 has roots \displaystyle 3 \pm \sqrt{5}, about 0.76 and 5.24; x = 3 gives -5.
(-\infty, \infty), every real number. The discriminant is 1 - 4 = -3 < 0, so there are no real roots, and x = 0 gives 1 > 0.
No solution. x^2 - 4x + 7 = (x - 2)^2 + 3, which is always at least 3.
(-\infty, -5) \cup (-5, \infty), every number except -5. x^2 + 10x + 25 = (x + 5)^2, which is 0 only at x = -5.
Only \displaystyle x = \frac{3}{2}. 4x^2 - 12x + 9 = (2x - 3)^2, never negative, and 0 only at \displaystyle x = \frac{3}{2}.
For 2 < t < 4 seconds. 30t - 5t^2 > 40 becomes t^2 - 6t + 8 < 0 after dividing by -5 and flipping, so (t - 2)(t - 4) < 0.
k < -4 or k > 4. The discriminant k^2 - 16 must be positive; k = 0 gives -16, so the middle stretch is out.
[-2, 4]. Need 8 + 2x - x^2 \ge 0, that is x^2 - 2x - 8 \le 0, so (x - 4)(x + 2) \le 0.
8 \le x \le 12. The other side is 20 - x, so x(20 - x) \ge 96, which becomes x^2 - 20x + 96 \le 0, that is (x - 8)(x - 12) \le 0.
x^2 - 3x - 4 > 0. The ends -1 and 4 are left out, so they are the roots of a strict inequality: (x + 1)(x - 4) > 0, which is positive outside the roots.
The quadratic’s graph is an unbroken curve, so to change sign it would have to pass through zero, and inside a stretch there is no root. So every number in a stretch gives the same sign as the one you tested.
Move everything to one side, so the inequality compares a quadratic with 0. Its solution can start or stop only where the quadratic’s sign changes or where it equals 0, and both happen only at roots. The set (0, 1) \cup (2, 3) starts or stops at four numbers, 0, 1, 2 and 3, and each of them would have to be a root. A quadratic has at most two.
x^2 - 9 = (x + 3)(x - 3) < 0 exactly when -3 < x < 3, by testing x = 0. And |x| < 3 says x is less than 3 away from 0, which is also -3 < x < 3.