Find where the expression is zero, then test each stretch between those points.
Solve for x
x² − x − 6 < 0
Factoring gives (x − 3)(x + 2) < 0, so the critical numbers are −2 and 3. The parabola opens upward, so it dips below zero only between its roots: −2 < x < 3. A is the answer to the opposite inequality, x² − x − 6 > 0. Test x = 0: 0 − 0 − 6 = −6, which is less than 0 ✓ — and 0 lies between −2 and 3.
Solve the equation first; the roots are where the sign can change.
Between two consecutive roots the expression cannot change sign — it would have to pass through zero to do so, and there are no other zeros. That is why one test point per interval settles the whole stretch.
x² − x − 6 < 0
zero on one side already
(x − 3)(x + 2) = 0
critical numbers −2 and 3
test −3, 0 and 4
one from each interval
−2 < x < 3
only the middle is negative
x² < x + 6 must become x² − x − 6 < 0 before you factor. Comparing factors against a non-zero number tells you nothing about their signs.
Useful as a check, but test a point anyway. It costs one substitution and catches a sign slip that this reasoning would not.
| Inequality | Interval | Endpoints |
|---|---|---|
| −2 < x < 3 | (−2, 3) | excluded, round brackets |
| −2 ≤ x ≤ 3 | [−2, 3] | included, square brackets |
| x < −2 or x > 3 | (−∞, −2) ∪ (3, ∞) | two pieces, joined by ∪ |
Infinity is not a number you can reach, so it is always written with a round bracket — even when the inequality uses ≤.
A bare square, a parabola that never touches the axis, and a word problem. The same four steps settle all three.
A bare square: x² ≤ 49
x² − 49 ≤ 0
zero on one side
(x − 7)(x + 7) ≤ 0
critical numbers −7 and 7
−7 ≤ x ≤ 7
test 0: 0 ≤ 49 ✓, so the middle
"x² ≤ 49, so x ≤ 7" looks finished, but it lets in x = −10, and (−10)² = 100 is not ≤ 49. A square root gives two numbers, ±7, so there are two critical numbers, not one.
No real roots: x² − 2x + 3 < 0
b² − 4ac = 4 − 12 = −8
negative: no critical numbers
test 0: 0 − 0 + 3 = 3
positive
no solution
the sign never changes
With no roots the whole number line is one interval, so one test point decides everything: the expression is always positive here, so < 0 has no solution and > 0 would hold for every x.
A word problem: a garden plot is 2 m longer than it is wide, and its area must be at most 48 m². What is the greatest whole-number width?
w(w + 2) ≤ 48
width w, length w + 2
w² + 2w − 48 ≤ 0
zero on one side
(w + 8)(w − 6) ≤ 0
−8 ≤ w ≤ 6
0 < w ≤ 6, so w = 6 m
a width must be positive
Check: 6 × 8 = 48 ≤ 48 ✓, while 7 × 9 = 63 is too big. The context throws away the negative half of the interval.
The same method; these are the places the answer looks different.
A negative leading coefficient: −x² + 5x − 4 ≥ 0
x² − 5x + 4 ≤ 0
multiply by −1 and flip the sign
(x − 1)(x − 4) ≤ 0
critical numbers 1 and 4
1 ≤ x ≤ 4
test 2: −4 + 10 − 4 = 2 ≥ 0 ✓
Different: multiplying by a negative reverses the inequality, exactly as in a linear one.
A double root: (x − 3)² > 0
(x − 3)² = 0 only at x = 3
one critical number
a square is positive everywhere else
test 0: 9 > 0 ✓
x ≠ 3
(−∞, 3) ∪ (3, ∞)
Different: the sign does not change at a double root, so the answer is every number but one.
Roots that do not factor: x² − 2x − 1 < 0
x = 2 ± √82
quadratic formula
x = 1 ± √2
√8 = 2√2
1 − √2 < x < 1 + √2
test 0: −1 < 0 ✓, the middle
Different: the critical numbers are irrational, but the sign chart works exactly the same.
y > x² − 4
a parabola boundary
draw y = x² − 4 dashed
> excludes the curve
test (0, 0): 0 > −4 ✓
true
shade the side containing the origin
inside the parabola
Same rules as a linear inequality: dashed for strict, solid for ≤ and ≥, and one test point decides the side. Only the boundary's shape has changed.
Periodical exams ask us to solve and pick the interval, or to pick one value that satisfies; entrance tests add a negative x², a bare x² < k or a length that must stay positive. The fastest route is roots first, then the parabola's shape to choose between "between" and "outside". The 10-second check: put x = 0 (or any easy number) into the original and see whether our answer includes it when it should.
Solve for x
x² − 16 ≥ 0
Factoring gives (x − 4)(x + 4) ≥ 0, so the critical numbers are ±4. The parabola opens upward, so it is at or above zero outside the roots: x ≤ −4 or x ≥ 4. Test x = 0: −16 ≥ 0 is false, so the middle is excluded — which rules out A. B forgets the negative branch: x = −5 gives 25 − 16 = 9 ≥ 0 ✓
Hint: test x = 0. If it fails, the solution is the two outer pieces, not the middle one.
