Quadratic and Rational Inequalities: Sign-Table Method.
An inequality like x(x+1)/(x−2) > 0 looks fierce, but solving it is closer to counting. Find where each factor becomes zero, cut the number line there, and in each piece count only signs: positive or negative. That is the sign-table method.
1. Compare with zero, then count signs
x(x+1)/(x−2) > 0 is true exactly when the signs of all the factors, top and bottom, multiply out to a positive whole: all three positive, or two negative and one positive. The factors vanish at −1, 0 and 2. These are the critical values, and they cut the number line into four intervals. Inside one interval no factor changes sign; a sign only changes when you cross a critical value.
Walk x along the line yourself below. In whichever interval x enters, each factor's sign is written in, and the last row gives the sign of the whole. Once every interval has been visited, the solution turns blue. Try changing the sign too: a zero of the denominator never fills in.
build a sign tablemove x
Signs change at critical values
Example 19 in the book is exactly this table. The sign is − for x < −1, + for −1 < x < 0, − for 0 < x < 2 and + for x > 2. We want > 0, so the solution set is {x : −1 < x < 0} ∪ {x : x > 2}. Write the steps out once by hand.
fill the signs
Example 19, step by step
Step 1. Find the critical values: x = 0, x + 1 = 0 gives x = −1, x − 2 = 0 gives x = 2. In order: −1, 0, 2.
Step 2. Sign of x on each interval: −, −, +, +. Sign of x + 1: −, +, +, +. Sign of x − 2: −, −, −, +.
Step 3. Multiply down each column: for x < −1, (−)(−)/(−) = −; for −1 < x < 0, (−)(+)/(−) = +; for 0 < x < 2, (+)(+)/(−) = −; for x > 2, (+)(+)/(+) = +.
Step 4. Keep the + intervals. Because the sign is strict, the critical values stay out: the solution is (−1, 0) ∪ (2, ∞).
2. Two fractions facing each other
Example 20 has a fraction on each side: (x+2)/(x+1) > (x−3)/(x−4). You may not cross-multiply, because you do not know the signs of the denominators. Bring the right-hand fraction across and use a common denominator. The numerator is (x+2)(x−4) − (x−3)(x+1) = (x² − 2x − 8) − (x² − 2x − 3) = −5. That leaves −5/((x+1)(x−4)) > 0, so (x+1)(x−4) < 0, and the solution is −1 < x < 4.
Board questions use the same move. In (2x+3)/(x−3) < (x+3)/(x−1) the numerator is x² + x + 6, whose discriminant 1 − 24 is negative, so it is always positive; what is left is (x−3)(x−1) < 0, solution 1 < x < 3. In (x − 3)/x < (x + 2)/(x + 3) the numerator is −2x − 9; multiplying by −1 flips the sign to (2x + 9)/[x(x + 3)] > 0, with critical values −9/2, −3, 0 and solution (−9/2, −3) ∪ (0, ∞). All three are worked below.
compare two fractionsbring to one side
Simplify the numerator first
When an absolute value appears, open it first. Example 21: |(x−3)/(2x+1)| > 3 means (x−3)/(2x+1) > 3 or < −3. The first gives −6/5 < x < −1/2, the second −1/2 < x < 0. Together the solution is −6/5 < x < 0 with x ≠ −1/2, since the denominator vanishes there. The set has supremum 0 and infimum −6/5.
Example 23(b) gives f/g = (x+2y−4)/(2x−y−3) > 0 and asks you to put x = y. The expression becomes (3x−4)/(x−3) > 0, with critical values 4/3 and 3, and the solution is x < 4/3 or x > 3. It is one of the choices in the sign table above.
3. The shortcut for quadratics
Example 24(b): factorise x² − 4x + 3 > 0 into (x−1)(x−3) > 0. The sign table gives + for x < 1, − for 1 < x < 3 and + for x > 3. So the solution is x < 1 or x > 3.
The picture is the same every time, so the book gives a shortcut. With a > b, the solution of (x−a)(x−b) ≤ 0 is b ≤ x ≤ a, between the roots. The solution of (x−a)(x−b) ≥ 0 is x ≥ a or x ≤ b, outside the roots. Move a, b and the sign below.
two rootsdrag a and b
Between, or outside?
Now use the shortcut. (x−4)(x−5) > 0 has solution x < 4 or x > 5; writing “and” here would leave no value at all. x² − 8x + 12 > 0 has solution (−∞, 2) ∪ (6, ∞). Multiply 5x − x² − 6 > 0 by −1 to get x² − 5x + 6 < 0, so 2 < x < 3. And x² + 2x − 3 < 0 has solution −3 < x < 1. Put the inequalities below into the right box.
pick the shapechip, then bin
Which shape of answer?
First make the x² coefficient positive, then look at the sign. Pick up a chip, then drop it in a box.
Right. Next one.
Not yet. If the x² coefficient is negative, multiply by −1 and the sign turns round.
All in place. Outside: x < 4 or x > 5; x < 2 or x > 6; x < −1 or x > 6. Between: 1 < x < 3; 1 ≤ x ≤ 2; 2 < x < 3.
4. A few more, from the boards
In (x−1)/(x² + 5x + 6) > 0, factorise the denominator: (x−1)/((x+2)(x+3)) > 0. The critical values are −3, −2, 1 and the signs from the left are −, +, −, +. The solution is (−3, −2) ∪ (1, ∞), and −3 and −2 are never included.
For L = {x ∈ ℝ : 2x² + 5x < 0}, x(2x+5) < 0 gives −5/2 < x < 0. Its midpoint is −5/4 and its half-length 5/4, so in absolute value it reads |x + 5/4| < 5/4. And f(x) = x² − x ≤ 0 means x(x − 1) ≤ 0, solution 0 ≤ x ≤ 1; the sign is ≤, so both ends are in.
5. Check yourself
Three questions in the exam's style. Critical values first, then signs.
check yourselfpick one answer
Question 1
Which is the solution of (x − 4)(x − 5) > 0?
x > 4 and x < 5 does not fit here; the right answer is x < 4 or x > 5. Both factors positive (x > 5) or both negative (x < 4).
Right. Both factors positive (x > 5) or both negative (x < 4).
x < 4 and x > 5 does not fit here; the right answer is x < 4 or x > 5. Both factors positive (x > 5) or both negative (x < 4).
x > 4 or x < 5 does not fit here; the right answer is x < 4 or x > 5. Both factors positive (x > 5) or both negative (x < 4).
check yourselfpick one answer
Question 2
Which is the solution set of x² − 8x + 12 > 0?
(−∞, 2] ∪ [6, ∞) does not fit here; the right answer is (−∞, 2) ∪ (6, ∞). (x − 2)(x − 6) > 0 ⇒ x < 2 or x > 6; the sign is strict so the ends are excluded.
[2, 6] does not fit here; the right answer is (−∞, 2) ∪ (6, ∞). (x − 2)(x − 6) > 0 ⇒ x < 2 or x > 6; the sign is strict so the ends are excluded.
Right. (x − 2)(x − 6) > 0 ⇒ x < 2 or x > 6; the sign is strict so the ends are excluded.
(2, 6) does not fit here; the right answer is (−∞, 2) ∪ (6, ∞). (x − 2)(x − 6) > 0 ⇒ x < 2 or x > 6; the sign is strict so the ends are excluded.
check yourselfpick one answer
Question 3
(x − 2)(x − 5) > 0 holds when,
i. x < 2 ii. 2 < x < 5 iii. x > 5
This combination does not match the true statements; the true ones are i, iii. For x < 2 and x > 5 the two factors have the same sign (i and iii true); for 2 < x < 5 the product is negative.
Right. For x < 2 and x > 5 the two factors have the same sign (i and iii true); for 2 < x < 5 the product is negative.
This combination does not match the true statements; the true ones are i, iii. For x < 2 and x > 5 the two factors have the same sign (i and iii true); for 2 < x < 5 the product is negative.
This combination does not match the true statements; the true ones are i, iii. For x < 2 and x > 5 the two factors have the same sign (i and iii true); for 2 < x < 5 the product is negative.
Finally, turn the cards over.
cut from the notetap to turn over
Eight cards
QuestionWhat is the solution of x(x+1)/(x−2) > 0?
Answer−1 < x < 0 or x > 2.
QuestionIn (x+2)/(x+1) > (x−3)/(x−4), what is the numerator after subtracting?
Answer−5, so (x+1)(x−4) < 0 and the solution is −1 < x < 4.
QuestionWhat is the solution of (3x−4)/(x−3) > 0?
Answerx < 4/3 or x > 3.
QuestionFor a > b, what is the solution of (x−a)(x−b) ≤ 0?
Answerb ≤ x ≤ a.
QuestionFor a > b, what is the solution of (x−a)(x−b) ≥ 0?
Answerx ≥ a or x ≤ b.
QuestionIs the solution of (x−4)(x−5) > 0 written with "and" or "or"?
Answer"Or": x < 4 or x > 5; "and" would give no value.
QuestionWhat is the solution of (2x+3)/(x−3) < (x+3)/(x−1)?
AnswerThe numerator x² + x + 6 is always positive, so (x−3)(x−1) < 0; solution 1 < x < 3.
QuestionWhat is the solution of (x − 1)/(x² + 5x + 6) > 0?