MCR 3U Functions Grade 11
QUADRATIC FUNCTIONS AND EQUATIONS
TEST 4
K/U
𝟏𝟑
APP
𝟏𝟏
TIPS
𝟕
Full solutions are required to receive full marks.
Knowledge and Understanding
1. Simplify: [4 marks]
a). √80
Solution
√80 = √16 × 5 = √16 × √5 = 4√5.
b). 6√10 × 10√5
Solution
6√10 × 10√5 = (6 ×10) (√10 × 5) = 60√50
= 60√25 × 2 = (60×5)√2
= 300√2
√40 − 2
c).
2
Solution
√40 − 2
2
√4×10 − 2
2
=
=
2√10 − 2
2
d). 6 − √−36
Solution
6 − √−36 = 6 – 6√−1 = 6 – 6i
= √10 – 1
COM
𝟐
2. Simplify:
√2 − √5
2 √5 + 5 √2
.
[2 marks]
Solution
√2 − √5
2√5 + 5√2
=2
=
=
=
√2 − √5
√5 + 5√2
2√5 − 5√2
√5 − 5√2
×2
√2 ∙ 2√5 − √2 ∙ 5√2 − √5 ∙ 2√5 + √5 ∙ 5√2
2
(2√5) − (5√2)
2
2√10 − 10 − 10 + 5√10
20− 50
7√10 − 20
− 30
=
20 − 7√10
30
.
3. Find the extremum (maximum or minimum) value and the value for x
when it occurs. [3 marks]
y = – 0.5x2 + 4x – 13
Solution 1
Since a = – 0.5 < 0, we have maximum.
𝑏
4
4
x=– =–
=–
= 4,
2𝑎
2(−0.5)
−1
ymax = y(4) = – 0.5(4 ) + 4(4) – 13 = – 8 + 16 – 13 = – 5.
Solution 2
y = – 0.5(x2 – 8x) – 13 = – 0.5(x2 – 8x + 16 – 16) – 13
= – 0.5[(x – 4)2 – 16] – 13
= – 0.5 (x – 4)2 + 8 – 13
= – 0.5 (x – 4)2 – 5
∴ ymax = y(4) = – 5.
2
4. Solve by factoring and check: 2z2 – 8 = 6z. [4 marks]
Solution
2z2 – 6z – 8 = 0 Divide by 2:
z2 – 3z – 4 = 0
(z – 4)(z + 1) = 0
z = 4 or z = – 1
Check:
1) z = 4
L.S. = 2(4)2 – 8 R.S. = 6(4)
= 32 – 8
= 24
= 24
L.S. = R.S.
2) z = – 1
L.S. = 2(– 1)2 – 8 R.S. = 6(– 1)
=6–8
= – 24
=–2
L.S. = R.S.
1
7
5. Solve using the quadratic formula: 𝑏 = 𝑏 2 − 1. [4 marks]
2
2
Solution
Multiply by 2: b = 7b2 – 2 ⟹ 7b2 – b – 2 = 0.
b=
1 ±√(−1)2 − 4(7)(−2)
b1 =
2(7)
1− √57
14
=
1 ±√1+ 56
≈ – 0.47, b2 =
14
1+ √57
14
=
1 ±√57
14
≈ 0.61
Applications
6. The function y = 4x2 – 4x + m has exactly one zero. Find the value of
𝑚.
[2 marks]
Solution
Since the quadratic trinomial 4x2 – 4x + m has exactly one zero, its
discriminant equals 0:
D = b2 – 4ac = (– 4)2 – 4(4)m = 0⟹ 16 – 16m = 0 ⟹ m = 1.
7. Three pieces of rod measures 20 cm, 41 cm, and 44 cm. If the same
amount is cut off each piece, the remaining lengths can be formed into a
right triangle. Find the cut off length. [4 marks]
Solution
Let x cm the cut off length, 0 < x < 20, then the new lengths of the rods
would be:
20 – x, 41 – x, 44 – x. By Pythagorean Theorem we get:
(20 – x)2 + (41 – x)2 = (44 – x)2
400 – 40x + x2 + 1681 – 82x + x2 = 1936 – 88x + x2.
x2 – 34x + 145 = 0
(x – 29)(x – 5) = 0
x – 29 = 0 or x – 5 = 0
x = 29 or x = 5.
Since x < 20, we get: x = 5 cm.
8. When a ball is thrown straight upward at a speed of 15 m/s, its height
in meters after t seconds is given by the function h (t) = 20t – 5t2.
[5 marks]
a). Find the maximum height of the ball to the nearest metre.
Solution
h (t) = – 5t2 + 20t = – 5 (t2 – 4t)
= – 5 (t2 – 4t + 4 – 4)
= – 5 [(t – 2)2 – 4] = – 5 (t – 2)2 + 20
∴ hmax = 20 when t = 2.
b). How long was the ball 12 m above the ground? Give the answer to
the nearest tenth of a second.
Solution
h (t) = 12
– 5t2 + 20t = 12,
5t2 – 20t + 12 = 0,
t=
20 ± √202 − 4(5)(12)
2(5)
=
20 ± √60
10
≈
20 ± 7.7
10
t1 ≈1.23
t2 ≈ 2.77
∆𝑡 = t2 – t1 ≈ 2.77 – 1.23 ≈ 1.5 s
The ball was 12 m above the ground for 1.5 s.
T.I.P.S.
9. If a bus travelled 10 km/h faster, it would take 2 h less to make a 315
km trip. Find the speed of the bus. [4 marks]
Solution
315
Let x km/h represent the speed of the bus then it takes the bus
h to
𝑥
make a 315 km trip. If the bus travelled 10 km/h faster, then its speed
315
would be (x + 10) km/h and it would take the bus
h to make a 315
𝑥+10
km trip. Since it would take 2 h less to make a 315 km trip, we get:
315
315
–
= 2 Multiply by x(x + 10):
𝑥
𝑥+10
315(x + 10) – 315x = 2x(x + 10)
315x + 3 150 – 315x = 2x2 + 20x
2x2 + 20x – 3150 = 0
x2 + 10x – 1575 = 0
(x + 45)(x – 35) = 0
x + 45 = 0 or x – 35 = 0
x = –45 or x = 35
Since x > 0, we get: x = 35.
The speed of the bus is 35 km/h
10. The numbers 2 + √3 and 2 – √3 are the roots of a quadratic
equation. Find this equation. [3 marks]
Solution
[x – (2 + √3)][x – (2 – √3)]= 0
[(x – 2) + √3][(x – 2) – √3]= 0
(x – 2)2 – (√3)2 = 0
x2 – 4x + 4 – 3 = 0
x2 – 4x + 1 = 0
Communications
11. Explain how complex numbers are different from real numbers.
Describe a real number system and use examples to illustrate your
explanations. [2 marks]
Solution
Real numbers are numbers that can be found on the number line. This
includes both the rational and irrational numbers. Rational numbers are
numbers that could be represented as fractions. For example,
2
3
and −
5
2
are rational numbers. Irrational numbers cannot be represented as
fractions. Examples of irrational numbers are 𝜋 and √2. You can add,
subtract, multiply and divide (not by 0) any two real numbers. You can
use real numbers to measure lengths of line segments. But you cannot
find square roots of negative real numbers. Thus, √−1 is undefined as a
real number. That's because you have no real number which square is –
1. However, you can take the square root of a negative number, but it
involves using a new number system to do it. In this system, called
complex numbers, √−1 exists and is called imaginary unit i. Thus, i =
√−1 and, hence, i2 = – 1. Any complex number z could be expressed in
the form z = x + iy, where a and b are real numbers.