Revision – Quadratic equations IB10
1.
(a)
Factorize the expression x2 − 25.
(b)
Factorize the expression x2 – 3x – 4.
(c)
Using your answer to part (b), or otherwise, solve the equation x2 – 3x – 4 = 0.
Working:
Answers:
(a) .....................................................
(b) .....................................................
(c) .....................................................
(Total 8 marks)
2.
(a)
Find the solution of the equation x2 – 5x – 24 = 0.
(b)
The equation ax2 – 9x – 30 = 0 has solution x = 5 and x = –2. Find the value of a.
Working:
Answers:
(a) ..............................................
(b) ..............................................
(Total 8 marks)
Revision IB10
Feb 2013
MaWa
3.
(a)
Solve the following equation for x
3(2x +1) − 2(3 − x) =13.
(2)
(b)
Factorize the expression x2 + 2x − 3.
(2)
(c)
Find the positive solution of the equation
x2 + 2x − 6 = 0.
(2)
Working:
Answers:
(a) ...................................................
(b) ...................................................
(c) ...................................................
(Total 6 marks)
Revision IB10
Feb 2013
MaWa
4.
The diagram below shows a path x m wide around a rectangular lawn which measures 10 m by
8 m.
xm
8m
LAWN
xm
10 m
(a)
Write down an expression in terms of x for the area of the path.
(b)
What is the width of the path when its area is 208 m2?
Working:
Answers:
(a) …………………………………………..
(b) …………………………………………..
(Total 4 marks)
Revision IB10
Feb 2013
MaWa
5.
Mrs Harvey wants to put a 50 m long fence around her rectangular garden. She only needs to
fence in 3 sides because the other side is alongside her house.
house
x
Diagram not to scale
garden
y
The width of the garden is denoted by x, and the length by y.
(a)
Write an expression for y in terms of x.
(b)
Write an expression for the area, A, of the garden, in terms of x.
(c)
If the area is 200 m2, find the dimensions of the garden.
Working:
Answers:
(a) ..................................................................
(b) ..................................................................
(c) ..................................................................
(Total 8 marks)
Revision IB10
Feb 2013
MaWa
6.
The length of one side of a rectangle is 2 cm longer than its width.
(a)
If the smaller side is x cm, find the perimeter of the rectangle in terms of x.
The perimeter of a square is equal to the perimeter of the rectangle in part (a).
(b)
Determine the length of each side of the square in terms of x.
The sum of the areas of the rectangle and the square is 2x2 + 4x +1 (cm2).
(c)
(i)
Given that this sum is 49 cm2, find x.
(ii)
Find the area of the square.
Working:
Answers:
(a) .................................................
(b) .................................................
(c)
(i).........................................
(ii)........................................
(Total 6 marks)
Revision IB10
Feb 2013
MaWa
7.
A swimming pool is to be built in the shape of a letter L. The shape is formed from two squares
with side dimensions x and
x as shown.
x
Diagram not to scale
x
x
x
(a)
Write down an expression for the area of the swimming pool surface.
(b)
The area A is to be 30 m2. Write a quadratic equation that expresses this information.
(c)
Find both the solutions of your equation in part (b).
(d)
Which of the solutions in part (c) is the correct value of x for the pool? State briefly why
you made this choice.
(Total 8 marks)
Markscheme for the Revision of Quadratic Equations IB10
1.
(a)
(x – 5)(x
5)
(b)
(x – 4)(x
1)
(c)
x=4
(A1)
x=–1
(A1)
(M1)(A1)(A1) (C3)
(M1)(A1)(A1) (
[8]
2.
(a)
(x – 8)(x + 3) = 0
x = 8, x = –3
(b)
METHOD 1
(x – 5)(x + 2) = 0
x2 – 3x – 10 = 0
3x2– 9x – 30 = 0
a=3
(M1)(M1)
(A1)(A1)(C2)(C2)
(M1)
(A1)
(A1)
(A1)
(C4)
METHOD 2
a(5)2 – 9(5) – 30 = 0
25a – 75 = 0
a=3
(M1)
(A2)
(A1)
(C4)
Revision IB10
Feb 2013
MaWa
(C2)
METHOD 3
a(–2)2 – 9(–2) – 30 = 0
4a – 12 = 0
a=3
(M1)
(A2)
(A1)
(C4)
[8]
3.
(a)
6x + 3 – 6 + 2x = 13
8x = 16
(M1)
x=2
(A1)
(C2)
(A1)(A1)
(C2)
(A2)
(C2)
(b)
(x + 3) (x – 1)
(c)
x = 1.64575..
x = 1.65
Note: If formula is used award (M1)(A1) for correct solution. If
graph is sketched award (M1)(A1) for correct solution.
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4.
(a)
(b)
A = 2(8x) + 2x(10 + 2x) or 2(10x) + 2x(8 + 2x)
or (10 + 2x)(8 + 2x) – 80
= 4x(x + 9) (or equivalent)
(M1)
(A1)
A = 4x(x + 9) = 208 (follow through from part (a))
x = 4 (or Width = 4)
(M1)
(A1)
[4]
5.
(a)
x + x + y = 50
y = 50 – 2x
(A1)
(A1) (C2)
(b)
A(x) = x(50 – 2x) or A(x) = 50x – 2x2
(c)
50x – 2x2 = 200
x2– 25x + 100 = 0
(x – 20)(x – 5) = 0
x = 20 or x = 5
(A2)(C2)
(M1)
(A1)
dimensions: 5 m × 40 m or
5m
40 m
(A2) or (C2)
OR
dimensions: 10 m × 20 m or
10 m
20 m
(A2) or (C2)(C4)
[8]
Revision IB10
Feb 2013
MaWa
6.
Unit penalty (UP) is applicable where indicated.
(a)
P (rectangle) = 2x + 2 (x + 2) = 4x + 4 cm
(A1)(C1)(UP)
(b)
Side of square = (4x + 4)/4 = x + 1 cm
(A1)(ft)(C1)(UP)
(c)
(i)
2x2 + 4x + 1 = 49 or equivalent
(M1)
(x + 6)(x – 4) = 0
x = – 6 and 4
(A1)
Note: award (A1) for the values or for correct factors
Choose x = 4
(A1)(ft)(C3)
Note: Award (A1)(ft) for choosing positive value.
(ii)
Area of square = 5 5 = 25 cm2
(A1)(ft)(C1)
Note: Follow through from both (b) and (c)(i).
(UP)
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7.
(a)
A = x2 + x or any equivalent unsimplified expression
Note: Award (A1) for each term.
(A1)(A1) (C2)
(b)
x2 + x = 30 or x2 + x – 30 = 0
Note: The answer must be an equation.
(C1)
(c)
(x – 5)(x + 6) = 0 or reasonable attempt to use formula.
(M1)(M1)
Note: Award (M1) for both signs wrong or one error in
quadratic formula (if used).
x = 5 or x = –6
(A1)(A1) (C4)
Note: Award (A2) d for x = 5 seen with no other working.
(d)
x = 5 because length must be positive (must have reason for the mark.)
(C1)
[8]
Revision IB10
Feb 2013
MaWa