Quadratic Inequalities Practice
Quadratic Inequalities Practice. Practice questions to deepen understanding of quadratic inequalities. Online math practice with full solutions and step-by-step explanations.
Quadratic inequalities practice — basic forms x²
🔢 Basic solution:
Solve: \(x^2 < 9\)
Step-by-step solution:
x² < 9
x² - 9 < 0
(x - 3)(x + 3) < 0
📊 Zero points: x = -3, x = 3
🔍 Check:
x = 0: 0² = 0 < 9 ✓
x = 4: 16 > 9 ✗
→ -3 < x < 3
🔢 Solve: \(x^2 \geq 16\)
Solution:
x² ≥ 16
x² - 16 ≥ 0
(x - 4)(x + 4) ≥ 0
Points: x = -4, x = 4
Check:
x = 0: 0 < 16 ✗
x = 5: 25 > 16 ✓
→ x ≤ -4 or x ≥ 4
🔢 Solve: \(x^2 - 5x + 6 < 0\)
Solution:
x² - 5x + 6 < 0
(x - 2)(x - 3) < 0
Zero points: 2, 3
Check (x = 2.5):
6.25 - 12.5 + 6 = -0.25 < 0 ✓
→ 2 < x < 3
🔢 Solve: \(x^2 + 3x - 4 \leq 0\)
Solution:
x² + 3x - 4 ≤ 0
(x + 4)(x - 1) ≤ 0
Points: -4, 1
Check (x = 0):
0 + 0 - 4 = -4 < 0 ✓
Endpoints are included.
→ -4 ≤ x ≤ 1
🔢 Solve: \(x^2 - 7x + 12 > 0\)
Solution:
x² - 7x + 12 > 0
(x - 3)(x - 4) > 0
Points: 3, 4
Check:
x = 0: 12 > 0 ✓
x = 3.5: negative ✗
x = 5: positive ✓
→ x < 3 or x > 4
🔢 Solve: \(2x^2 - 8 < 0\)
Solution:
2x² - 8 < 0
2x² < 8
x² < 4
(x - 2)(x + 2) < 0
Points: -2, 2
→ -2 < x < 2
🔢 Solve: \(x^2 + 2x - 8 \geq 0\)
Solution:
x² + 2x - 8 ≥ 0
(x + 4)(x - 2) ≥ 0
Points: -4, 2
Check (x = 0): -8 < 0 ✗
Endpoints are included.
→ x ≤ -4 or x ≥ 2
🔢 Solve: \(x^2 - 1 \leq 0\)
Solution:
x² - 1 ≤ 0
(x - 1)(x + 1) ≤ 0
Points: -1, 1
Check (x = 0): -1 ≤ 0 ✓
Endpoints included.
→ -1 ≤ x ≤ 1
🔢 Solve: \(x^2 + 4x + 4 > 0\)
Solution:
x² + 4x + 4 = (x + 2)²
(x + 2)² > 0
A square is always ≥ 0 and equals 0 only at x = -2.
→ x ≠ -2
🔢 Solve: \(x^2 - 4x \geq 0\)
Solution:
x² - 4x ≥ 0
x(x - 4) ≥ 0
Points: 0, 4
Check:
x = 2 → negative
x = -1 → positive ✓
x = 5 → positive ✓
→ x ≤ 0 or x ≥ 4
Solve: \(x^2 + x < 0\)
Solution:
x² + x < 0
x(x + 1) < 0
Points: -1, 0
Check (x = -0.5):
positive × negative = negative
→ -1 < x < 0
Solve: \(x^2 - 2x - 3 \leq 0\)
Solution:
x² - 2x - 3 ≤ 0
(x - 3)(x + 1) ≤ 0
Points: -1, 3
Check (x = 0):
-3 ≤ 0
→ -1 ≤ x ≤ 3
Solve: \(x^2 + 6x + 9 \geq 0\)
Solution:
x² + 6x + 9 = (x + 3)²
A square is always ≥ 0 for all x.
→ All real numbers
Solve: \(x^2 - 9 > 0\)
Solution:
x² - 9 > 0
(x - 3)(x + 3) > 0
→ x < -3 or x > 3
Solve: \(x^2 + 5x + 6 \geq 0\)
Solution:
x² + 5x + 6 ≥ 0
(x + 3)(x + 2) ≥ 0
→ x ≤ -3 or x ≥ -2
📐 Solve: \(x^2 - 4x + 3 < 0\)
Solution using the discriminant:
Δ = 16 - 12 = 4
x₁,₂ = (4 ± 2)/2
x₁ = 1, x₂ = 3
(x - 1)(x - 3) < 0
→ 1 < x < 3
📐 Solve: \(x^2 + 5x + 6 \geq 0\)
Solution:
Δ = 25 - 24 = 1
x₁,₂ = (-5 ± 1)/2
x₁ = -3, x₂ = -2
(x + 3)(x + 2) ≥ 0
→ x ≤ -3 or x ≥ -2
📐 Solve: \(x^2 - 2x - 3 > 0\)
\(x^2 - 2x - 3 = 0\)
Discriminant: \(\Delta = (-2)^2 - 4(1)(-3) = 4 + 12 = 16\)
\(x_{1,2} = \frac{2 \pm 4}{2}\)
\(x_1 = -1, \quad x_2 = 3\)
\(x^2 - 2x - 3 = (x+1)(x-3)\)
The inequality becomes:
\((x+1)(x-3) > 0\)
The parabola opens upward (\(a=1>0\)) ⌣
Positive (above the x-axis):
outside the roots!
\(x < -1\) or \(x > 3\) ✓
📐 Solve: \(2x^2 - 5x + 2 \leq 0\)
\(2x^2 - 5x + 2 = 0\)
Discriminant: \(\Delta = 25 - 16 = 9\)
\(x_{1,2} = \frac{5 \pm 3}{4}\)
\(x_1 = 0.5, \quad x_2 = 2\)
The parabola opens upward (\(a=2>0\)) ⌣
Negative or zero (below or on the x-axis):
between the roots!
\(0.5 \leq x \leq 2\) ✓
The sign is \(\leq\) (and not \(<\))
so the roots themselves are included!
Closed interval \([0.5, 2]\)
📐 Solve: \(x^2 + 4x + 4 < 0\)
\(x^2 + 4x + 4 = (x+2)^2\)
This is a perfect square!
The inequality becomes:
\((x+2)^2 < 0\)
But:
A square is always \(\geq 0\)!
It is impossible for a square to be negative
There is no \(x\) for which \((x+2)^2 < 0\)
No solution (empty set ∅)
Even at \(x=-2\): \(0 < 0\) false!
🔢 Solve: \(x^2 < x\)
Solution:
x² < x
x² - x < 0
x(x - 1) < 0
Points: 0, 1
→ 0 < x < 1
🔢 Solve: \(x^2 + 1 > 2x\)
Solution:
x² + 1 > 2x
x² - 2x + 1 > 0
(x - 1)² > 0
True for every x ≠ 1!
→ all x except 1
🔢 Solve: \((x - 1)(x + 2) < 0\)
Direct solution!
(x - 1)(x + 2) < 0
Already factored!
Points: 1, -2
→ -2 < x < 1
🔢 Solve: \(x^2 \geq 2x + 3\)
Solution:
x² ≥ 2x + 3
x² - 2x - 3 ≥ 0
(x - 3)(x + 1) ≥ 0
→ x ≤ -1 or x ≥ 3
🔢 Solve: \(4 - x^2 > 0\)
Solution:
4 - x² > 0
4 > x²
x² < 4
Points: -2, 2
→ -2 < x < 2
🔢 Solve: \(x^2 - 10x + 25 \geq 0\)
Solution:
x² - 10x + 25 ≥ 0
(x - 5)² ≥ 0
A square is always ≥ 0!
→ all ℝ
🔢 Solve: \(3x^2 - 12 < 0\)
Solution:
3x² - 12 < 0
3x² < 12
x² < 4
→ -2 < x < 2
🔢 Solve: \(x^2 + 6x + 9 > 0\)
Solution:
x² + 6x + 9 > 0
(x + 3)² > 0
True for every x ≠ -3!
→ all x except -3
🔢 Solve: \(x(x - 4) \leq 0\)
Direct solution:
x(x - 4) ≤ 0
Already factored!
Points: 0, 4
→ 0 ≤ x ≤ 4
🎯 Summary:
Solve: \(x^2 + x - 6 < 0\)
Full solution:
x² + x - 6 < 0
Δ = 1 + 24 = 25
x₁,₂ = (-1 ± 5)/2
x₁ = -3, x₂ = 2
(x + 3)(x - 2) < 0
Check (x = 0): -6 < 0 ✓
→ -3 < x < 2