The answer is a stretch of the number line, not a point.
Solve for x
x² − 4 < 0
x < 2 alone is not the answer
The roots are ±2, and an upward parabola dips below zero only between its roots: −2 < x < 2. B is the trap the card warns about — square-rooting both sides quietly loses the lower bound.
Zero is already on the right, so go straight to factoring.
x² − 4 < 0
x · x and 2 · 2
(x − 2)(x + 2) < 0
x² < 4 does not give x < 2 on its own. That answer would let x = −10 through, and (−10)² is 100.
A difference of squares always splits into two brackets.
These are the only places the sign can change.
x − 2 = 0
x = 2
x + 2 = 0
x = −2
the line is now cut into three zones
At the boundaries the product is 0, and 0 is not < 0.
| Zone | Try | (x − 2) | (x + 2) | Product |
|---|---|---|---|---|
| x < −2 | −3 | neg | neg | POSITIVE |
| −2 < x < 2 | 0 | neg | pos | NEGATIVE |
| x > 2 | 3 | pos | pos | POSITIVE |
We want the product to be NEGATIVE.
try x = 0: (−2)(2) = −4
negative, so the middle zone works
A negative times a negative is positive, so only the zone with one of each sign can be negative.
( −2 , 2 )
because it is <, not ≤
One stretch, written as a single sandwich.
Below zero means below the x-axis.
between the roots
outside the roots
Flip the sign to > 0 and you take the two outside pieces instead.
Same method: factor, mark the roots, pick the zone.
(x − 3)(x + 2) ≤ 0
roots 3 and −2
try x = 0: (−3)(2) = −6
negative, so the middle zone
−2 ≤ x ≤ 3
≤ keeps the ends: closed dots
| Upward parabola (a > 0) | The answer is |
|---|---|
| < 0 or ≤ 0 | between the roots |
| > 0 or ≥ 0 | outside the roots: x below the smaller or above the larger |
If a is negative, multiply through by −1 first and flip the sign: −x² + x + 6 ≥ 0 becomes x² − x − 6 ≤ 0, the inequality we just solved.
A perfect square touches the axis once; a negative discriminant never touches it.
| Inequality | Solution |
|---|---|
| (x − 4)² > 0 | every x except 4 |
| (x − 4)² ≥ 0 | every real x |
| (x − 4)² < 0 | no solution |
| (x − 4)² ≤ 0 | only x = 4 |
| x² + 2x + 5 > 0 (D = 4 − 20 < 0) | every real x |
x² + kx + 9 > 0 for every x
upward and never touching the axis
D = k² − 36 < 0
no real roots
−6 < k < 6
(x − 4)² < 0 has no solution at all, and (x − 4)² > 0 loses only x = 4. Factoring and testing zones still works, but the zones on both sides of 4 come out positive.
Same four moves; each case adds one thing to watch.
A negative x²: −x² + 4x + 12 < 0
x² − 4x − 12 > 0
× −1, and the sign flips
(x − 6)(x + 2) > 0
roots 6 and −2
x < −2 or x > 6
> 0: outside the roots
Test x = 0 in the original: 0 + 0 + 12 = 12, not < 0, so the middle zone really is out.
Multiplying by −1 without turning < into > gives −2 < x < 6 — exactly the zones that fail.
Not zero on the right: (x + 1)(x − 3) ≥ 5
x² − 2x − 3 − 5 ≥ 0
expand, then subtract 5
(x − 4)(x + 2) ≥ 0
x² − 2x − 8
x ≤ −2 or x ≥ 4
outside, ends included
The brackets given are useless: −1 and 3 are not the boundaries. Only a zero on the right lets the signs of the brackets decide.
Roots that are surds: x² − 2x − 1 < 0
x = 2 ± √82 = 1 ± √2
it will not factor; use the formula
1 − √2 < x < 1 + √2
< 0: between the roots
about −0.41 < x < 2.41
integers inside: 0, 1, 2
The boundaries come from the formula instead of brackets; the zone rule is unchanged.
The ends of the interval are the roots.
roots −3 and 4
the ends of the interval
(x + 3)(x − 4) < 0
between the roots, so < 0
x² − x − 12 < 0
expand
b = −1, c = −12
Inequality items test the zone and the ends: between or outside, open or closed, or how many integers fit. Ten-second check: try one value inside your answer and one outside — and an endpoint, to settle ≤ against <.
Only the middle zone passes, and the edges are excluded.
Solve the inequality
x² − 9 ≥ 0
The roots are ±3, and an upward parabola sits at or above zero outside them: x ≤ −3 or x ≥ 3. C is the inside region, where the expression is negative. Flip the inequality and you swap which side you keep.
Hint: the sign points the other way, so you keep the outside zones. The answer comes in two pieces, and ≥ makes the dots closed.
