Quadratic Inequalities (Grade 11) — Worksheet & Memo

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Grade 11 · Equations & Inequalities

Quadratic Inequalities (Grade 11)

MARKING GUIDELINE

Marks

Duration

55 minutes

Questions

Name:

Class:

Date:

Mark

/ 34

Instructions and Information

Answer ALL the questions in this question paper.
Answer QUESTION 1 by circling the letter (A–D) in the answer grid at the end of that section.
Show ALL calculations clearly.
Show all units where applicable.
Number the answers correctly according to the numbering system used in this question paper.
A non-programmable calculator may be used, unless stated otherwise.
Write neatly and legibly.

Question 1

[10 MARKS]

Four options are given as possible answers to the following questions. Choose the correct answer and circle the letter (A–D) in the grid at the end of this section. If you want to change your choice, put a cross through the wrong letter and circle your new choice.

1.1
Solve for x: x² − 4 > 0
(1)
A)x < −2 or x > 2✓
B)−2 < x < 2
C)x > 2 only
D)x > −2
Answer: A — The parabola lies above the x-axis OUTSIDE its roots −2 and 2.
- B — that is where the parabola is BELOW the axis
- C — the branch x < −2 works as well
- D — between the roots the expression is negative, so that set is wrong
1.2
Solve for x: x² − 4 < 0
(1)
A)−2 < x < 2✓
B)x < −2 or x > 2
C)x < 2
D)x < 4
Answer: A — The parabola dips below the x-axis BETWEEN its roots.
- B — that is where the parabola is ABOVE the axis
- C — x = −5 satisfies x < 2 but gives 21, which is positive
- D — square-rooted one side only and never used the roots
1.3
Solve for x: x(x − 4) + 2(4 − x) ≤ 0
(1)
A)2 ≤ x ≤ 4✓
B)x ≤ 2 or x ≥ 4
C)−2 ≤ x ≤ −4
D)0 ≤ x ≤ 4
Answer: A — The expression simplifies to x² − 6x + 8 = (x − 2)(x − 4), which is ≤ 0 between its roots.
- B — that is where the expression is POSITIVE
- C — the roots are +2 and +4, not negative
- D — read factors off the original brackets without simplifying first
1.4
Solve for x: (x + 1)(x − 5) ≥ 0
(1)
A)x ≤ −1 or x ≥ 5✓
B)−1 ≤ x ≤ 5
C)x ≥ −1
D)x ≥ 5
Answer: A — A product of two factors is positive OUTSIDE the roots.
- B — that is where the product is NEGATIVE
- C — x = 0 lies in that set but gives (1)(−5) = −5
- D — the branch x ≤ −1 works as well
1.5
Solve for x: x² ≥ 9
(1)
A)x ≤ −3 or x ≥ 3✓
B)x ≥ 3
C)−3 ≤ x ≤ 3
D)x ≥ −3
Answer: A — x² − 9 ≥ 0 holds outside the roots ±3.
- B — x = −4 works too: (−4)² = 16, which is ≥ 9
- C — that is where x² ≤ 9
- D — x = 0 lies in that set, but 0 is not ≥ 9
1.6
The FIRST step in solving a quadratic inequality is to …
(1)
A)move everything to one side so it is compared with 0, then factorise✓
B)divide both sides by x
C)take the square root of both sides
D)reverse the inequality sign
Answer: A — The sign of the expression can only be read off once it is a product compared with zero.
- B — dividing by x is not allowed — x may be zero or negative
- C — square-rooting loses the negative branch of the solution
- D — the sign changes only when you multiply or divide by a negative
1.7
Solve for x: x² + x − 6 ≤ 0
(1)
A)−3 ≤ x ≤ 2✓
B)x ≤ −3 or x ≥ 2
C)−2 ≤ x ≤ 3
D)x ≤ 2
Answer: A — (x + 3)(x − 2) ≤ 0 holds between the roots −3 and 2.
- B — that is where the expression is POSITIVE
- C — the signs of the two roots have been swapped
- D — x = −10 satisfies x ≤ 2 but gives 84, which is positive
1.8
If a quadratic has NO real roots and a > 0, then ax² + bx + c > 0 for …
(1)
A)all real values of x✓
B)no values of x
C)x > 0 only
D)x < 0 only
Answer: A — The parabola opens upward and never reaches the axis, so it lies entirely above it.
- B — the whole graph is above the axis, so every x works
- C — the graph is above the axis for negative x as well
- D — the graph is above the axis for positive x as well
1.9
Solve for x: x² < 0
(1)
A)there is no real value of x✓
B)x < 0
C)x = 0
D)all real x
Answer: A — A square is never negative.
- B — (−3)² = 9, which is positive
- C — 0² = 0, which is not LESS than 0
- D — no square is negative, so no x works
1.10
Solve for x: (x − 3)² > 0
(1)
A)every real x except x = 3✓
B)x > 3
C)all real x
D)x = 3
Answer: A — A square is positive everywhere except where it is zero, and it is zero at x = 3.
- B — x = 0 gives 9, which is greater than 0, so x < 3 works too
- C — at x = 3 the value is 0, which is not GREATER than 0
- D — at x = 3 the value is exactly 0

Answer grid — marking guideline

1.1	A	B	C	D
1.2	A	B	C	D
1.3	A	B	C	D
1.4	A	B	C	D
1.5	A	B	C	D
1.6	A	B	C	D
1.7	A	B	C	D
1.8	A	B	C	D
1.9	A	B	C	D
1.10	A	B	C	D

Question 2

[12 MARKS]

Solve for x. Show the critical values and give your answers in interval notation where asked.

2.1
2x² − 5x + 3 < 0
(4)
(2x − 3)(x − 1) = 0 gives critical values x = 32 and x = 1 (2)
The parabola opens upwards, and < 0 means BELOW the axis — between the roots (1)
1 < x < 32 (1)
2.2
x² − x − 6 ≥ 0
(4)
(x − 3)(x + 2) = 0 gives x = 3 and x = −2 (2)
≥ 0 means ON or ABOVE the axis — outside the roots (1)
x ≤ −2 or x ≥ 3 (1)
2.3
−x² + 4 > 0
(4)
Multiply by −1 and REVERSE: x² − 4 < 0 (2)
Critical values ±2 (1)
−2 < x < 2 (1)

Question 3

[12 MARKS]

Answer the questions below.

3.1
Solve for x: x + 1x − 2 ≥ 0
(5)
The critical values are x = −1 and x = 2 (2)
x = 2 must be EXCLUDED — the fraction is undefined there (1)
Test each interval (1)
x ≤ −1 or x > 2 (1)
3.2
Write your answer to the previous part in interval notation.
(2)
(−∞ ; −1] ∪ (2 ; ∞) (2)
3.3
For which values of x is the graph of y = x² − x − 6 BELOW the graph of y = 2x − 6? Show all your working.
(5)
x² − x − 6 < 2x − 6 (1)
x² − 3x < 0 (1)
x(x − 3) < 0 (1)
0 < x < 3 (2)

TOTAL: 34 marks

This question paper consists of 3 questions.