Solve the quadratic inequality:
x2 - 4x - 21 < 0
x2 - 4x - 21 < 0
Worked Solution & Answer
Critical values from (x + 3)(x - 7) = 0 are x = -3 and x = 7, so -3 < x < 7. B1
Answer: -3 < x < 7
Quadratic inequalities — Notes & Worked Examples Show / Hide
Explanation:
Never solve a quadratic inequality like a linear equation: always rearrange so one side is 0 (with a positive x2 term), factorise to find the critical values (roots), and sketch a quick U-shaped parabola crossing the x-axis at those roots.
• If the quadratic is < 0 (below the x-axis), the solution is a single connected region between the roots: a < x < b.
• If the quadratic is > 0 (above the x-axis), the solution is two separate regions outside the roots: x < a or x > b.
Example 1: Solve x2 − 4x − 12 < 0
Solution:
Factorise: (x − 6)(x + 2) < 0
Critical values (roots): x = −2 and x = 6
U-shaped curve is below the x-axis (< 0) between the roots:
Answer: −2 < x < 6
Example 2: Solve 3x2 ≥ 10x + 8
Solution:
Rearrange & factorise: 3x2 − 10x − 8 = (3x + 2)(x − 4) ≥ 0
Critical values: x = −23 and x = 4
Curve is on or above x-axis (≥ 0) outside the roots:
Answer: x ≤ −23 or x ≥ 4
Solve the quadratic inequality:
x2 - 2x - 8 < 0
x2 - 2x - 8 < 0
Worked Solution & Answer
Critical values from (x + 2)(x - 4) = 0 are x = -2 and x = 4, so -2 < x < 4. B1
Answer: -2 < x < 4
Solve 2x2 + 3x - 2 > 0
Worked Solution & Answer
Factorise: (2x – 1)(x + 2) > 0. Critical values are x = –2 and x = 1/2. Since the parabola opens upwards, 2x² + 3x – 2 > 0 for x < –2 or x > 1/2. M2 A1
Answer: x < -2 or x > 12
Here is a sketch of the curve y = x2 + 4x - 12
Work out the values of x for which x2 + 4x - 12 < 0
Give your answer as an inequality.
Give your answer as an inequality.
Answer
Worked Solution & Answer
Solve x2 + 4x - 12 = 0: M1
(x + 6)(x - 2) = 0 ⇒ x = -6 or x = 2 M1
The curve is a parabola opening upwards, so y < 0 between the two roots:
-6 < x < 2 A1
(x + 6)(x - 2) = 0 ⇒ x = -6 or x = 2 M1
The curve is a parabola opening upwards, so y < 0 between the two roots:
-6 < x < 2 A1
Answer: Answer -6 < x < 2
Solve the inequality.
x2 - 100 ≥ 0
x2 - 100 ≥ 0
Worked Solution & Answer
(x + 10)(x − 10) ⩾ 0 M1
⇒ critical values −10 and 10 M1
⇒ x ⩽ −10 or x ⩾ 10. A1
⇒ critical values −10 and 10 M1
⇒ x ⩽ −10 or x ⩾ 10. A1
Answer: x ≤ -10,\ x ≥ 10
(a) Solve x² – 5x – 6 < 0. [2 marks]
25 (b) Show the solution on the number line. [1 mark]
25 (b) Show the solution on the number line. [1 mark]
(a) Solve x2 - 5x - 6 < 0 [2 marks]
Answer
(b) Show the solution to x2 - 5x - 6 < 0 on the number line. [1 mark]
Worked Solution & Answer
(a)
(x – 6)(x + 1) < 0 M1
⇒ roots at x = –1 and x = 6, so –1 < x < 6 A1
⇒ roots at x = –1 and x = 6, so –1 < x < 6 A1
Answer (a): Answer -1 < x < 6
(b)
Open circles at –1 and 6 joined by a line B1
Solve 2x2 > 12 - 5x
Answer
Worked Solution & Answer
Rearrange: 2x² + 5x – 12 > 0. M1
Factorise: (2x – 3)(x + 4) > 0. M1
Critical values: x = 1.5 and x = –4. M1
Solution: x < –4 or x > 1.5. A1
Factorise: (2x – 3)(x + 4) > 0. M1
Critical values: x = 1.5 and x = –4. M1
Solution: x < –4 or x > 1.5. A1
Answer: Answer x < -4 or x > 1.5
Solve the inequality.
x2 - 5x - 6 ≤ 0
x2 - 5x - 6 ≤ 0
Worked Solution & Answer
Factorise: (x − 6)(x + 1) ⩽ 0. M2
Critical values are x = −1 and x = 6. M1
Since the parabola opens upwards, x² − 5x − 6 ⩽ 0 for −1 ⩽ x ⩽ 6. A1
Critical values are x = −1 and x = 6. M1
Since the parabola opens upwards, x² − 5x − 6 ⩽ 0 for −1 ⩽ x ⩽ 6. A1
Answer: -1 ≤ x ≤ 6
Solve the inequality 4x2 - 5x - 6 > 0
Worked Solution & Answer
(4x + 3)(x – 2) > 0 M2
Critical values x = –3/4 and x = 2 M1
Outside region: x < –3/4 or x > 2 A1
Critical values x = –3/4 and x = 2 M1
Outside region: x < –3/4 or x > 2 A1
Answer: x < -0.75, x > 2
(a) Solve x² + 2x > 6x + 5. (3)
(b) Represent your solution set to part (a) on the number line. (1)
(b) Represent your solution set to part (a) on the number line. (1)
(a) Solve x2 + 2x > 6x + 5 [3 marks]
(b) Represent your solution set to part (a) on the number line below. [1 mark]
Worked Solution & Answer
(a)
x² – 4x – 5 > 0 M1
⇒ (x – 5)(x + 1) > 0 M1
⇒ x < –1 or x > 5 A1
⇒ (x – 5)(x + 1) > 0 M1
⇒ x < –1 or x > 5 A1
Answer (a): x < -1, x > 5
(b)
Open circles at –1 and 5 with arrows pointing left from –1 and right from 5
Answer (b): Open circles at –1 and 5 with arrows pointing left from –1 and right from 5