C. Polynomial and Rational Functions · Rational Functions and Equations

Solving Rational Inequalities Algebraically with Sign Charts

C4.1 C4.2About 75 minutes

Learning goals

I can explain why multiplying both sides of an inequality by an expression like x − 3 can give a wrong answer.
I can rewrite a rational inequality with 0 on one side and a single fraction on the other.
I can use the critical values and a sign chart (or a table of intervals) to solve the inequality.
I can decide which boundary values to include and check my answer with test values or a graph.

Words to know

critical value
A zero of the numerator or of the denominator of the single fraction. The sign can only change at these values.

sign chart
A number line, divided at the critical values, showing the sign of each factor and of the whole expression in each interval.

test value
A number chosen inside an interval to find the sign of the expression on the whole interval.

reversing the inequality
Multiplying or dividing both sides by a negative number flips the inequality sign, for example −2x < 6 becomes x > −3.

Why we need a new method

Graphs show the solution of an inequality clearly, but not always exactly. Algebra gives exact boundaries. There is one trap, and it is a big one.

The trap. Try to solve 2/(x − 3) < 1 by multiplying both sides by x − 3. You get 2 < x − 3, so x > 5. But test x = 0: 2/(0 − 3) = −2/3, and −2/3 < 1 is true! So x = 0 is a solution that the method missed.

What went wrong? When x < 3, x − 3 is negative, and multiplying an inequality by a negative number reverses it. Since you do not know the sign of x − 3 in advance, multiplying by it is unsafe.

The graph of y = 2/(x − 3) is below y = 1 on the whole left branch (x < 3) and on the right branch after x = 5. The open circle at (5, 1) shows the boundary is not included for '<'. The careless method only finds x > 5.

The safe method

Move everything to one side so the other side is 0.
Combine into a single fraction and factor the numerator and the denominator.
Find the critical values: zeros of the numerator and zeros of the denominator.
Make a sign chart: divide the number line at the critical values and find the sign in each interval (with test values, or by multiplying the signs of the factors).
Choose the intervals that match the inequality. Include zeros of the numerator for ≤ or ≥; never include zeros of the denominator.

This is the same sign-chart idea you use for polynomial inequalities in Polynomial Functions and Equations, with one addition: the zeros of the denominator are critical values too.

Worked example

Example 1: Already in the right form

Solve (x − 1)/(x + 4) ≥ 0.

Critical values: 1 (numerator zero) and −4 (denominator zero).
Test x = −5: (−6)/(−1) = 6 > 0. Test x = 0: −1/4 < 0. Test x = 2: 1/6 > 0.
We want ≥ 0: the positive intervals, plus x = 1 (where it equals 0). x = −4 is excluded.

x < −4 or x ≥ 1, or (−∞, −4) ∪ [1, ∞).

Worked example

Example 2: Fixing the trap

Solve 2/(x − 3) < 1 correctly.

Move everything to one side: 2/(x − 3) − 1 < 0.
Single fraction: [2 − (x − 3)]/(x − 3) = (5 − x)/(x − 3) < 0.
Critical values: 5 and 3.
Signs: x = 0 gives 5/(−3) < 0; x = 4 gives 1/1 > 0; x = 6 gives (−1)/3 < 0.
We want < 0, so take the negative intervals and exclude both critical values.

x < 3 or x > 5. The careless method lost the whole interval x < 3.

The factor signs multiply to give − | + | −. The shaded intervals are the solution of (5 − x)/(x − 3) < 0. Both circles are open: 3 is undefined and 5 gives 0, which is not less than 0.

Worked example

Example 3: Three critical values

Solve (x + 2)/(x² − 9) > 0.

Factor: (x + 2)/((x + 3)(x − 3)). Critical values: −3, −2, 3.
Use the factor signs (see the chart): − | + | − | +.
Check with test values: x = −4: (−2)/7 < 0. x = −2.5: (−0.5)/(−2.75) > 0. x = 0: 2/(−9) < 0. x = 4: 6/7 > 0 ✓.
We want > 0: the second and fourth intervals, with no endpoints.

−3 < x < −2 or x > 3, or (−3, −2) ∪ (3, ∞).

With three factors, multiply the three signs in each column. An even number of minus signs gives +, an odd number gives −.

Worked example

Example 4: Fractions on both sides

Solve 3/(x + 1) ≤ 2/(x − 1).

One side 0: 3/(x + 1) − 2/(x − 1) ≤ 0.
Common denominator: [3(x − 1) − 2(x + 1)]/((x + 1)(x − 1)) = (x − 5)/((x + 1)(x − 1)) ≤ 0.
Critical values: −1, 1, 5.
Tests: x = −2: (−7)/((−1)(−3)) = −7/3 < 0 ✓. x = 0: (−5)/((1)(−1)) = 5 > 0. x = 2: (−3)/((3)(1)) = −1 < 0 ✓. x = 6: 1/((7)(5)) > 0.
Include x = 5 (value 0); exclude −1 and 1.

x < −1 or 1 < x ≤ 5, or (−∞, −1) ∪ (1, 5].

Worked example

Example 5: When multiplying is safe

Solve 1/(x² + 1) < 1/2.

Both denominators are always positive (x² + 1 ≥ 1 and 2 > 0), so multiplying both sides by 2(x² + 1) keeps the direction.
2 < x² + 1, so x² > 1.
x² > 1 means x < −1 or x > 1.

x < −1 or x > 1. Multiplying is fine only when you know the sign of what you multiply by.

Just for fun

Multiplying both sides of an inequality by x − 3 without checking its sign is like heading out in a snowstorm without checking the forecast: sometimes you arrive fine, and sometimes you end up in the ditch with half your solutions missing.

Hands-on activity

Coin sign charts

You need: about 20 coins (or small squares of paper with + on one side and − on the other), a sheet of paper and a pencil.

Draw a long number line and mark the critical values −3, −2 and 3 for (x + 2)/((x + 3)(x − 3)). This makes four intervals.
Make three rows, one for each factor: x + 2, x + 3 and x − 3.
In each interval, place a coin heads up if that factor is positive there and tails up if it is negative. (Test with a number in the interval if you are unsure.)
Count the tails in each column. An even number of tails means the whole expression is positive; odd means negative.
Use your coins to solve (x + 2)/((x + 3)(x − 3)) > 0, then ≤ 0. Remember which critical values can be included.
Make up your own rational inequality for a family member and check their coin chart.

What to look for: columns, left to right: tails-tails-tails (3 tails, −), tails-heads-tails (2 tails, +), heads-heads-tails (1 tail, −), heads-heads-heads (0 tails, +). So > 0 on (−3, −2) ∪ (3, ∞), and ≤ 0 on (−∞, −3) ∪ [−2, 3). Only −2 can be included, because −3 and 3 make the denominator 0.

Your turn

Solve (x + 5)/(x − 2) > 0.

Solve (3 − x)/(x + 1) ≥ 0.

Solve 4/(x + 2) > 2.

Solve (x − 1)/(x² − 4) ≤ 0.

Why is it wrong to solve 4/(x + 2) > 2 by multiplying both sides by x + 2?

Solve x/(x − 1) < 2.

More practice: warm-up

List the critical values of (x − 4)/(x + 6).

Is (x − 4)/(x + 6) positive or negative at x = 0?

For (x − 4)/(x + 6) ≥ 0, which critical value can be included?

Solve x/(x − 2) < 0.

Solve 1/(x + 3) > 0.

More practice: on your own

Solve (2x + 1)/(x − 4) ≤ 0.

Solve 1/(x − 1) ≥ 1/(x + 1).

Solve 5/(x − 2) ≤ 3.

Solve (x + 1)/(x − 3) > 2.

Solve 6/(x² + 2) ≥ 1.

Challenge

Solve (x² − 1)/(x² − 4) < 0.

Solve 1/x < x.

Solve 2/(x + 1) − 1/(x − 2) ≥ 0.

Check your understanding

Quiz

Why is it risky to multiply both sides of 2/(x − 3) < 1 by x − 3?

What is the solution of (x − 1)/(x + 4) ≥ 0?

Which inequality is equivalent to 2/(x − 3) < 1?

What are the critical values of (x + 2)/(x² − 9)?

What is the solution of 3/(x + 1) ≤ 2/(x − 1)?

On a sign chart, how is a critical value that makes the denominator 0 shown?

For parents and teachers

Big idea (C4.1, C4.2): to solve a rational inequality exactly, get 0 on one side, write a single factored fraction, mark the critical values (zeros of the top and of the bottom), and use a sign chart. Never multiply both sides by an expression whose sign you do not know.

How to help: The coin activity makes sign charts concrete. Ask your teen to check one number inside and one outside each solution interval with a calculator.

Common difficulties:

Multiplying by the denominator and losing part of the solution.
Subtraction errors when combining into a single fraction (use brackets: 2 − (x − 3)).
Including a value that makes the denominator 0.
Forgetting critical values from the denominator in the sign chart.