Inequalities
An inequality compares two expressions with ${<}$, ${>}$, ${\le}$, or ${\ge}$. A solution is a number that makes the inequality true. Most inequalities have infinitely many solutions, so we write the answer as an interval (see real numbers) and draw it on a number line.
Rules for inequalities
We solve inequalities much like equations, with one big difference:
- Adding or subtracting the same number on both sides keeps the inequality sign the same.
- Multiplying or dividing both sides by a positive number keeps the sign the same.
- Multiplying or dividing both sides by a negative number reverses the sign: ${<}$ becomes ${>}$, and ${\le}$ becomes ${\ge}$.
Why the reversal? Start with ${2<5}$, which is true. Multiply both sides by ${-1}$: ${-2}$ and ${-5}$. Now ${-2}$ is the bigger one, so ${-2>-5}$. Multiplying by a negative number flips the order on the number line.
Linear inequalities
Example 1: Solve ${3-2x\le 11}$. Write the answer in interval notation, and graph it.
Solution:
Subtract ${3}$ from both sides:
${-2x\le 8}$
Divide both sides by ${-2}$. The divisor is negative, so the sign reverses:
${x\ge -4}$
In interval notation, the answer is ${[-4,\infty)}$.
Compound inequalities
A compound inequality joins two inequalities with "and" or "or".
- "And" means both must be true. We often write it as one three-part inequality, like ${-5<2x+3\le 7}$. To solve it, do the same step to all three parts.
- "Or" means at least one must be true. The answer is the union ${\cup}$ of the two solution sets: all the numbers in either one.
Example 2: Solve ${-5<2x+3\le 7}$.
Solution:
Subtract ${3}$ from all three parts. Then divide all three parts by ${2}$:
$\begin{align*}-8&<2x\le 4\\-4&<x\le 2\end{align*}$
In interval notation, the answer is ${(-4,2]}$.
Absolute value inequalities
Remember that ${|A|}$ is the distance from ${A}$ to ${0}$. Let ${k>0}$.
- ${|A|<k}$ means ${A}$ is less than ${k}$ units from ${0}$. So ${A}$ is between ${-k}$ and ${k}$:
${|A|<k}$ means ${-k<A<k}$. - ${|A|>k}$ means ${A}$ is more than ${k}$ units from ${0}$. So ${A}$ is to the left of ${-k}$ or to the right of ${k}$:
${|A|>k}$ means ${A<-k}$ or ${A>k}$.
The same rules hold with ${\le}$ and ${\ge}$.
Example 3: Solve each inequality.
(a) ${|x-3|<2}$
(b) ${|2x+1|\ge 5}$
Solution:
(a) Write it as a three-part inequality. Then add ${3}$ to all three parts:
$\begin{align*}-2&<x-3<2\\1&<x<5\end{align*}$
The answer is ${(1,5)}$. Since ${|x-3|}$ is the distance from ${x}$ to ${3}$, these are the numbers less than ${2}$ units from ${3}$.
(b) Split it into two inequalities joined by "or":
${2x+1\le -5}$ or ${2x+1\ge 5}$
Solve each one:
${2x\le -6}$ or ${2x\ge 4}$
${x\le -3}$ or ${x\ge 2}$
The answer is ${(-\infty,-3]\cup[2,\infty)}$.
Polynomial inequalities
For an inequality like ${x^2-x-6>0}$, we use a sign chart. The idea: a polynomial can change sign only where it equals ${0}$. Between those points, its sign stays the same.
- Move every term to one side, so the other side is ${0}$.
- Factor, and find where each factor is ${0}$. These are the critical numbers.
- The critical numbers split the number line into intervals. Test one number in each interval to find the sign there.
- Choose the intervals with the sign you need. Include the critical numbers only if the sign is ${\le}$ or ${\ge}$.
Example 4: Solve ${x^2-x-6>0}$.
Solution:
Factor the left side:
${(x-3)(x+2)>0}$
The critical numbers are ${3}$ and ${-2}$. Test one number in each interval:
| Interval | Test ${x}$ | Sign of ${(x-3)(x+2)}$ |
|---|---|---|
| ${(-\infty,-2)}$ | ${-3}$ | ${(-)(-)=+}$ |
| ${(-2,3)}$ | ${0}$ | ${(-)(+)=-}$ |
| ${(3,\infty)}$ | ${4}$ | ${(+)(+)=+}$ |
We need the product to be positive (${>0}$). The critical numbers make it ${0}$, so they are not included:
${(-\infty,-2)\cup(3,\infty)}$
Example 5: Solve ${x^3\ge 4x}$.
Solution:
Do not divide by ${x}$: its sign is unknown. Move every term to one side, and factor:
$\begin{align*}x^3-4x&\ge 0\\x(x+2)(x-2)&\ge 0\end{align*}$
The critical numbers are ${-2}$, ${0}$, and ${2}$. Find the sign of each factor in each interval:
| Interval | Sign of ${x(x+2)(x-2)}$ |
|---|---|
| ${(-\infty,-2)}$ | ${(-)(-)(-)=-}$ |
| ${(-2,0)}$ | ${(-)(+)(-)=+}$ |
| ${(0,2)}$ | ${(+)(+)(-)=-}$ |
| ${(2,\infty)}$ | ${(+)(+)(+)=+}$ |
We need ${\ge 0}$. So take the positive intervals, and include the critical numbers:
${[-2,0]\cup[2,\infty)}$
Rational inequalities
The same method works for fractions. But now the critical numbers include the zeros of the denominator too. Those numbers are never included in the answer, because the fraction is undefined there.
Watch out: do not multiply both sides by the denominator. Its sign is unknown, so you would not know whether to reverse the inequality.
Example 6: Solve ${\dfrac{x+1}{x-2}\le 0}$.
Solution:
The numerator is ${0}$ at ${x=-1}$. The denominator is ${0}$ at ${x=2}$. These are the critical numbers.
| Interval | Sign of ${\dfrac{x+1}{x-2}}$ |
|---|---|
| ${(-\infty,-1)}$ | ${\dfrac{-}{-}=+}$ |
| ${(-1,2)}$ | ${\dfrac{+}{-}=-}$ |
| ${(2,\infty)}$ | ${\dfrac{+}{+}=+}$ |
We need ${\le 0}$. The fraction is negative on ${(-1,2)}$. It equals ${0}$ at ${x=-1}$, so include ${-1}$. It is undefined at ${x=2}$, so leave ${2}$ out:
${[-1,2)}$
Example 7: Solve ${\dfrac{3}{x-1}\ge 2}$.
Solution:
First get ${0}$ on the right side. Subtract ${2}$, and combine into one fraction with the common denominator ${x-1}$:
$\begin{align*}&\dfrac{3}{x-1}-2\\&=\dfrac{3-2(x-1)}{x-1}\\&=\dfrac{5-2x}{x-1}\end{align*}$
So we need ${\dfrac{5-2x}{x-1}\ge 0}$. The critical numbers are ${\dfrac{5}{2}}$ (from the numerator) and ${1}$ (from the denominator).
| Interval | Sign of ${\dfrac{5-2x}{x-1}}$ |
|---|---|
| ${(-\infty,1)}$ | ${\dfrac{+}{-}=-}$ |
| ${\left(1,\frac{5}{2}\right)}$ | ${\dfrac{+}{+}=+}$ |
| ${\left(\frac{5}{2},\infty\right)}$ | ${\dfrac{-}{+}=-}$ |
We need ${\ge 0}$. Include ${\dfrac{5}{2}}$, where the fraction is ${0}$, but not ${1}$:
${\left(1,\dfrac{5}{2}\right]}$
An application
Example 8: A ball is thrown straight up from the ground at ${19.6}$ m/s. Its height after ${t}$ seconds is
${h=-4.9t^2+19.6t}$ meters.
During what time is the ball higher than ${14.7}$ m?
Solution:
The ball goes up, slows down, and comes back down. So it passes the height ${14.7}$ m twice: once going up and once coming down. In between, it is higher.
We need ${h>14.7}$:
${-4.9t^2+19.6t>14.7}$
Move every term to the left side:
${-4.9t^2+19.6t-14.7>0}$
Divide every term by ${-4.9}$. The divisor is negative, so the sign reverses:
${t^2-4t+3<0}$
Factor:
${(t-1)(t-3)<0}$
The product is negative between the critical numbers ${1}$ and ${3}$. So the ball is higher than ${14.7}$ m when ${1<t<3}$: from ${1}$ second to ${3}$ seconds after the throw.
Summary
- Multiplying or dividing an inequality by a negative number reverses the sign.
- Solve a three-part inequality by doing the same step to all three parts.
- ${|A|<k}$ means ${-k<A<k}$. ${|A|>k}$ means ${A<-k}$ or ${A>k}$.
- Polynomial inequalities: get ${0}$ on one side, factor, and use a sign chart.
- Rational inequalities: the zeros of the denominator are critical numbers too, and they are never included. Do not multiply by the denominator.