Solving Quadratic Equations: Every Method & Type — Worksheet & Memo

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Grade 11 · Equations

Solving Quadratic Equations: Every Method & Type

MARKING GUIDELINE

Marks

Duration

55 minutes

Questions

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Class:

Date:

Mark

/ 35

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 − 5x + 6 = 0
(1)
A)x = 2 or x = 3✓
B)x = −2 or x = −3
C)x = 1 or x = 6
D)x = 5 or x = 6
Answer: A — (x − 2)(x − 3) = 0, since 2 × 3 = 6 and 2 + 3 = 5.
- B — both factors are NEGATIVE, so the roots come out positive
- C — 1 × 6 = 6, but 1 + 6 = 7, not 5
- D — read off b and c instead of factorising
1.2
Solve for x: x(2x + 6) = 0
(1)
A)x = 0 or x = −3✓
B)x = −3 only
C)x = 0 or x = 3
D)x = −6 only
Answer: A — A product is zero when either factor is zero: x = 0, or 2x + 6 = 0.
- B — left out the root x = 0
- C — 2x + 6 = 0 gives x = −3, not +3
- D — did not divide the 6 by the 2, and dropped the x = 0 root
1.3
Solve for x, correct to TWO decimal places: 4x² = 9 − 3x
(1)
A)x = 1,17 or x = −1,92✓
B)x = −1,17 or x = 1,92
C)x = 1,50 or x = −1,50
D)x = 12,37 or x = −12,37
Answer: A — 4x² + 3x − 9 = 0, so x = (−3 ± √153) ÷ 8.
- B — the formula starts with −b, which is −3, so the signs are the wrong way round
- C — solved 4x² = 9 and ignored the −3x term altogether
- D — gave ±√153 without subtracting b or dividing by 2a
1.4
The quadratic formula is x = …
(1)
A)−b ± √(b² − 4ac)2a✓
B)b ± √(b² − 4ac)2a
C)−b ± √(b² + 4ac)2a
D)−b ± √(b² − 4ac)a
Answer: A — That is the formula as it appears on the information sheet.
- B — the formula begins with −b, not +b
- C — the discriminant is b² − 4ac, with a MINUS
- D — the denominator is 2a, not a
1.5
Solve for x: x² = 16
(1)
A)x = 4 or x = −4✓
B)x = 4 only
C)x = 8
D)x = 256
Answer: A — Both 4² and (−4)² equal 16.
- B — a square root has TWO values
- C — halved 16 instead of square-rooting it
- D — squared 16 instead of square-rooting it
1.6
Completing the square turns x² + 6x = 7 into …
(1)
A)(x + 3)² = 16✓
B)(x + 3)² = 7
C)(x + 6)² = 43
D)(x + 3)² = 9
Answer: A — Half of 6 is 3, so add 3² = 9 to BOTH sides: 7 + 9 = 16.
- B — added the 9 to the left-hand side only
- C — used 6 inside the bracket instead of half of 6
- D — wrote the 9 on the right instead of 7 + 9
1.7
Hence the solutions of x² + 6x = 7 are …
(1)
A)x = 1 or x = −7✓
B)x = 1 only
C)x = 4 or x = −4
D)x = 7 or x = −1
Answer: A — x + 3 = ±4, so x = 1 or x = −7.
- B — a square root has TWO values
- C — forgot to subtract the 3 from both answers
- D — made a sign slip on both roots
1.8
A quadratic equation has …
(1)
A)at most two real roots✓
B)exactly two real roots
C)exactly one real root
D)at least two real roots
Answer: A — A negative discriminant gives no real roots, and a zero discriminant gives exactly one.
- B — a negative discriminant gives no real roots at all
- C — that happens only when the discriminant is zero
- D — there may be none, so 'at least' is wrong
1.9
Solve for x: 2x² − 8 = 0
(1)
A)x = 2 or x = −2✓
B)x = 4 or x = −4
C)x = 2 only
D)x = √8
Answer: A — 2x² = 8 gives x² = 4, so x = ±2.
- B — forgot to divide by the 2 before square-rooting
- C — a square root has TWO values
- D — square-rooted the 8 without dividing by the 2 first
1.10
Two consecutive POSITIVE integers have a product of 42. The smaller one is …
(1)
A)6✓
B)7
C)21
D)42
Answer: A — x(x + 1) = 42 gives x² + x − 42 = (x + 7)(x − 6) = 0, and the positive root is 6.
- B — that is the LARGER of the two integers
- C — halved 42 instead of solving the equation
- D — read the product as though it were one of the integers

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. Where a calculator is needed, give your answers correct to TWO decimal places.

2.1
2x² − 5x + 3 = 0
(3)
(2x − 3)(x − 1) = 0 (2)
x = 32 or x = 1 (1)
2.2
x = x + 12
(5)
Square both sides: x² = x + 12 (1)
x² − x − 12 = 0 (1)
(x − 4)(x + 3) = 0 (1)
Check both in the ORIGINAL equation (1)
x = −3 gives 9 = 3 ≠ −3, so it is rejected: x = 4 (1)
2.3
3x² + 5x − 1 = 0, correct to TWO decimal places
(4)
x = −5 ± 25 + 126 (2)
= −5 ± 376 (1)
x = 0,18 or x = −1,85 (1)

Question 3

[13 MARKS]

Answer the questions below.

3.1
Solve for x and y simultaneously: 5x − y = 2 and 2x² − 3y = 4
(6)
From the linear equation: y = 5x − 2 (1)
2x² − 3(5x − 2) = 4 (1)
2x² − 15x + 2 = 0 (1)
x = 15 ± 2094 (1)
x = 7,36 gives y = 34,82 (1)
x = 0,14 gives y = -1,32 (1)
3.2
Determine the value(s) of k for which x² + (k + 2)x + 9 = 0 has EQUAL roots.
(4)
Equal roots means Δ = 0 (1)
(k + 2)² − 36 = 0 (2)
k = 4 or k = −8 (1)
3.3
Explain why the roots of a quadratic can be checked by substituting them back into the ORIGINAL equation, and why this matters when a surd equation is solved.
(3)
Squaring both sides can introduce a root that does not satisfy the original (2); only substitution into the original shows which one is valid (1)

TOTAL: 35 marks

This question paper consists of 3 questions.