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7.4D Quadratic Formula: Fluency and Exam Drill

EDEXCEL 1MA1 · Calculator allowed · about 85 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

For the equation 3x² + 5x - 2 = 0, write down the values of a, b and c that you would substitute into the quadratic formula.

(Total for Question 1 is 2 marks)

Write down the quadratic formula used to solve an equation of the form ax² + bx + c = 0.

(Total for Question 2 is 1 mark)

Solve x² + 5x - 3 = 0 using the quadratic formula. Give your answers correct to 2 decimal places.

(Total for Question 3 is 3 marks)

Solve 2x² + 5x - 3 = 0 using the quadratic formula. Give your answers correct to 2 decimal places.

(Total for Question 4 is 3 marks)

Calculate the value of the discriminant, b² - 4ac, for the equation 2x² - 5x + 1 = 0.

(Total for Question 5 is 2 marks)

A quadratic equation has discriminant b² - 4ac = -20. State how many real solutions the equation has, giving a reason for your answer.

(Total for Question 6 is 2 marks)

Solve x² - 4x - 7 = 0 using the quadratic formula. Give your answers correct to 2 decimal places.

(Total for Question 7 is 3 marks)

Solve 4x² + 4x - 15 = 0 using the quadratic formula. Give your answers as exact decimals.

(Total for Question 8 is 3 marks)

Solve x² + 6x + 2 = 0 using the quadratic formula. Give your answers correct to 3 significant figures.

(Total for Question 9 is 3 marks)

Solve 5x² - 7x - 1 = 0 using the quadratic formula. Give your answers correct to 2 decimal places.

(Total for Question 10 is 3 marks)

Solve 2x² - 3x - 6 = 0 using the quadratic formula. Give your answers correct to 2 decimal places.

(Total for Question 11 is 3 marks)

Solve x² + 7x + 4 = 0 using the quadratic formula. Give your answers correct to 2 decimal places.

(Total for Question 12 is 3 marks)

A number x satisfies x² - 5x = 9. Rearrange this equation into the form ax² + bx + c = 0, then use the quadratic formula to find the two possible values of x, giving your answers correct to 2 decimal places.

(Total for Question 13 is 4 marks)

A number x satisfies 2x² + 3x = 7. Rearrange this equation into the form ax² + bx + c = 0, then use the quadratic formula to find the two possible values of x, giving your answers correct to 2 decimal places.

(Total for Question 14 is 4 marks)

Two whole numbers differ by 3. The product of the two numbers is 88. By forming and solving a quadratic equation, find the two possible pairs of numbers.

(Total for Question 15 is 4 marks)

A stallholder at a craft market in Leeds models her daily profit, P pounds, from selling x greetings cards using P = -2x² + 40x - 150.

(a)Show that when P = 30, the equation can be simplified to x² - 20x + 90 = 0.(2)

(b)Solve x² - 20x + 90 = 0 using the quadratic formula, giving your answers correct to 1 decimal place.(3)

(c)Hence state the two numbers of greetings cards, to the nearest whole card, for which the daily profit is 30 pounds.(1)

(Total for Question 16 is 6 marks)

Two students solve x² - 6x + 4 = 0 using the quadratic formula.

(a)Use the quadratic formula to solve x² - 6x + 4 = 0, giving your answers correct to 2 decimal places.(3)

(b)Verify one of your solutions by substituting it back into x² - 6x + 4, and explain why the result is not exactly zero.(2)

(Total for Question 17 is 5 marks)

Solve x² - 10x + 13 = 0. Give your answers in the form p ± q*√3, where p and q are integers.

(Total for Question 18 is 3 marks)

In a badminton tournament, every player plays every other player exactly once. If there are x players, the total number of matches played is (x(x - 1))/2. There are 66 matches played in total. Form an equation in x and use the quadratic formula to find the number of players, explaining why the other solution to your equation is not valid.

(Total for Question 19 is 4 marks)

A group of x students are going on a school trip. The total cost of hiring a coach is 240 pounds, shared equally between the students. If 4 more students join the trip, the cost per student decreases by 2 pounds. Form an equation in x and solve it using the quadratic formula to find the original number of students.

(Total for Question 20 is 4 marks)

A cyclist travels 40 km at a steady speed of x km/h. On her return journey, a strong headwind reduces her speed by 2 km/h, and the return journey takes 1/3 of an hour (20 minutes) longer than the outward journey. Form an equation in x and solve it using the quadratic formula to find her outward speed, giving your answer correct to 1 decimal place.

(Total for Question 21 is 4 marks)

Solve the equation 3/(x + 1) + 2/(x - 2) = 1, where x is not equal to -1 and x is not equal to 2, giving your answers correct to 2 decimal places.

(Total for Question 22 is 5 marks)

Mark scheme · 7.4D Quadratic Formula: Fluency and Exam Drill

Question 1

B1 a = 3 and c = -2, both correct
B1 b = 5
Answer: a = 3, b = 5, c = -2

Question 2

B1 x = (-b ± √b² - 4ac) / (2a) oe, fully correct
Answer: x = (-b ± √b² - 4ac) / (2a)

Question 3

M1 correct substitution with a = 1, b = 5, c = -3, e.g. x = (-5 ± √5² - 4(1)(-3)) / (2(1))
A1 x = 0.54 (awrt 0.54)
A1 x = -5.54 (awrt -5.54)
Answer: x = 0.54 or x = -5.54

Question 4

M1 correct substitution with a = 2, b = 5, c = -3
A1 x = 0.5 cao
A1 x = -3 cao
Answer: x = 0.5 or x = -3

Question 5

M1 (-5)² - 4(2)(1) seen, i.e. 25 - 8
A1 17 cao
Answer: 17

Question 6

B1 0 (no real solutions)
B1 reason: the discriminant is negative, so √b² - 4ac is not a real number
Answer: 0 real solutions, since the discriminant is negative

Question 7

M1 correct substitution with a = 1, b = -4, c = -7
A1 x = 5.32 (awrt 5.32)
A1 x = -1.32 (awrt -1.32)
Answer: x = 5.32 or x = -1.32

Question 8

M1 correct substitution with a = 4, b = 4, c = -15
A1 x = 1.5 oe (e.g. 3/2), cao
A1 x = -2.5 oe (e.g. -5/2), cao
Answer: x = 1.5 or x = -2.5

Question 9

M1 correct substitution with a = 1, b = 6, c = 2
A1 x = -0.354 (awrt -0.354)
A1 x = -5.65 (awrt -5.65)
Answer: x = -0.354 or x = -5.65

Question 10

M1 correct substitution with a = 5, b = -7, c = -1
A1 x = 1.53 (awrt 1.53)
A1 x = -0.13 (awrt -0.13)
Answer: x = 1.53 or x = -0.13

Question 11

M1 correct substitution with a = 2, b = -3, c = -6
A1 x = 2.64 (awrt 2.64)
A1 x = -1.14 (awrt -1.14)
Answer: x = 2.64 or x = -1.14

Question 12

M1 correct substitution with a = 1, b = 7, c = 4
A1 x = -0.63 (awrt -0.63)
A1 x = -6.37 (awrt -6.37)
Answer: x = -0.63 or x = -6.37

Question 13

M1 rearranges to x² - 5x - 9 = 0 oe
M1 correct substitution with a = 1, b = -5, c = -9 (ft)
A1 x = 6.41 (awrt 6.41)
A1 x = -1.41 (awrt -1.41)
Answer: x = 6.41 or x = -1.41

Question 14

M1 rearranges to 2x² + 3x - 7 = 0 oe
M1 correct substitution with a = 2, b = 3, c = -7 (ft)
A1 x = 1.27 (awrt 1.27)
A1 x = -2.77 (awrt -2.77)
Answer: x = 1.27 or x = -2.77

Question 15

M1 forms the equation x(x + 3) = 88 oe, e.g. x² + 3x - 88 = 0
M1 correct substitution into the quadratic formula with a = 1, b = 3, c = -88 (ft)
A1 x = 8 or x = -11, both cao
A1 states both pairs: 8 and 11, or -11 and -8
Answer: 8 and 11, or -11 and -8

Question 16

(a) M1 sets -2x² + 40x - 150 = 30 and rearranges to -2x² + 40x - 180 = 0 oe
(a) A1 divides through by -2 to reach x² - 20x + 90 = 0 cso (all steps shown)
(a) Answer: x² - 20x + 90 = 0 (shown)
(b) M1 correct substitution with a = 1, b = -20, c = 90 (ft)
(b) A1 x = 13.2 (awrt 13.2)
(b) A1 x = 6.8 (awrt 6.8)
(b) Answer: x = 13.2 or x = 6.8
(c) B1 7 cards and 13 cards, both correct (ft from part b)
(c) Answer: 7 cards or 13 cards

Question 17

(a) M1 correct substitution with a = 1, b = -6, c = 4
(a) A1 x = 5.24 (awrt 5.24)
(a) A1 x = 0.76 (awrt 0.76)
(a) Answer: x = 5.24 or x = 0.76
(b) M1 substitutes x = 5.24 (or 0.76) into x² - 6x + 4, e.g. (5.24)² - 6(5.24) + 4
(b) A1 obtains a value close to 0 (e.g. 0.02), and explains this is due to rounding the exact root to 2 decimal places
(b) Answer: (5.24)² - 6(5.24) + 4 = 0.02 (awrt), close to zero; the small difference is caused by rounding the exact surd root to 2 decimal places

Question 18

M1 correct substitution with a = 1, b = -10, c = 13, leading to √48 seen
M1 simplifies √48 to 4*√3
A1 x = 5 + 2*√3 or x = 5 - 2*√3, both, fully simplified oe
Answer: x = 5 + 2*√3 or x = 5 - 2*√3

Question 19

M1 forms x(x - 1) = 132 oe, e.g. x² - x - 132 = 0
M1 correct substitution into the quadratic formula with a = 1, b = -1, c = -132 (ft)
A1 x = 12 cao
B1 rejects x = -11, with reason that the number of players cannot be negative
Answer: 12 players

Question 20

M1 forms 240/x - 240/(x + 4) = 2 oe
M1 multiplies through by x(x + 4) and rearranges to x² + 4x - 480 = 0 oe (unsimplified 2x² + 8x - 960 = 0 accepted at this stage)
M1 correct substitution into the quadratic formula with a = 1, b = 4, c = -480 (ft)
A1 x = 20 cao, rejecting x = -24 since the number of students cannot be negative
Answer: 20 students

Question 21

M1 forms 40/(x - 2) - 40/x = 1/3 oe
M1 multiplies through by 3x(x - 2) and rearranges to x² - 2x - 240 = 0 oe
M1 correct substitution into the quadratic formula with a = 1, b = -2, c = -240 (ft)
A1 x = 16.5 (awrt 16.5), rejecting the negative root since speed cannot be negative
Answer: 16.5 km/h

Question 22

M1 multiplies both sides by (x + 1)(x - 2) to clear the fractions
M1 correctly expands both sides: 3(x - 2) + 2(x + 1) = (x + 1)(x - 2)
dM1 rearranges to the 3-term quadratic x² - 6x + 2 = 0 (dependent on the previous method mark)
A1 x = 5.65 (awrt 5.65)
A1 x = 0.35 (awrt 0.35)
Answer: x = 5.65 or x = 0.35

Quadratic Formula: Fluency and Exam Drill - Worksheets, Questions and Revision