Expand and simplify expressions

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This document provides explanations and solved examples on expanding and simplifying expressions, finding values of x in equations, solving inequalities, factorising expressions, and finding the dimensions of a book.

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DIP1003 SEMESTER 1,2019
ASSIGNMENT 2
MARCH 26, 2019
Submitted by:
Timo Weiss

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1. Expand and simplify the following expressions:
a) (πŸ’π’™ + π’š)𝟐 βˆ’ πŸ‘π’™(π’š + πŸπ’™)
Sol.
Using (π‘Ž + 𝑏)2 = π‘Ž2 + 𝑏2 + 2π‘Žπ‘
(4π‘₯ + 𝑦)2 βˆ’ 3π‘₯(𝑦 + 2π‘₯)
= ((4π‘₯)2 + 𝑦2 + 2(4π‘₯)(𝑦)) βˆ’ 3π‘₯(𝑦 + 2π‘₯)
= 16π‘₯ 2 + 𝑦2 + 8π‘₯𝑦 βˆ’ 3π‘₯𝑦 βˆ’ 6π‘₯ 2
= 10π‘₯ 2 + 𝑦2 + 5π‘₯𝑦
b)
(π’™πŸ’π’šπŸŽπ’›πŸ”)
𝟏
𝟐(𝒙
βˆ’πŸ‘
𝟐 π’š
πŸ”
𝟐)
βˆ’πŸ
π’™πŸ’π’šπŸπ’›πŸ‘
π’™βˆ’πŸ’π’šπ’›πŸ‘
Sol.
(π‘₯4𝑦0𝑧6)1
2(π‘₯βˆ’3
2 𝑦6
2)βˆ’2π‘₯4𝑦2𝑧3
π‘₯βˆ’4𝑦𝑧3
= (π‘₯2𝑧3)π‘₯4𝑦2𝑧3
(π‘₯βˆ’3𝑦6)(π‘₯βˆ’4𝑦𝑧3)
= π‘₯6𝑦2𝑧6
π‘₯βˆ’7𝑦7𝑧3
= π‘₯13 π‘¦βˆ’5𝑧3
= π‘₯13 𝑧3
𝑦5
2. Fid the values of x in the following equation using the rule that if 𝒂 βˆ— 𝒃 = 𝟎 then 𝒂 =
𝟎 𝒐𝒓 𝒃 = 𝟎 . Do not expand the equation or use the quadratic formula.
(πŸ“π’™ + πŸ”) (πŸ‘π’™ βˆ’ 𝟐
πŸ‘) = 𝟎
Sol.
πΈπ‘–π‘‘β„Žπ‘’π‘Ÿ (5π‘₯ + 6) = 0 π‘œπ‘Ÿ (3π‘₯ βˆ’ 2
3) = 0
5π‘₯ + 6 = 0
π‘₯ = βˆ’6
5
(3π‘₯ βˆ’ 2
3) = 0
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3π‘₯ = 2
3
π‘₯ = 2
9
π‘₯ = βˆ’6
5 π‘œπ‘Ÿ 2
9
3. Solve the following equation for π’Ž.
Hint: π’‚πŸ βˆ’ π’ƒπŸ = (𝒂 βˆ’ 𝒃)(𝒂 + 𝒃)
(πŸπ’Ž)𝟐 βˆ’ πŸ’
π’ŽπŸ + 𝟏 = πŸπ’Ž + 𝟐
π’Ž + 𝟐
Sol.
(2π‘š)2 βˆ’ 4
π‘š2 + 1 = 2π‘š + 2
π‘š + 2
= (2π‘š)2 βˆ’ 22
π‘š2 + 1 = 2π‘š + 2
π‘š + 2
= (2π‘š + 2)(2π‘š βˆ’ 2)
π‘š2 + 1 = 2π‘š + 2
π‘š + 2
= 4(π‘š + 1)(π‘š βˆ’ 1)
π‘š2 + 1 = 2(π‘š + 1)
π‘š + 2
= 2(π‘š βˆ’ 1)
π‘š2 + 1 = 1
π‘š + 2
2(π‘š βˆ’ 1)(π‘š + 2) = π‘š2 + 1
2(π‘š2 + 2π‘š βˆ’ π‘š βˆ’ 2) = π‘š2 + 1
2π‘š2 + π‘š βˆ’ 2 = π‘š 2 + 1
= π‘š2 + π‘š βˆ’ 3
4. Factorise the following expressions:
a) πŸ‘πŸ”π’™πŸ βˆ’ πŸ–πŸ
Sol.
36π‘₯2 βˆ’ 81
= (6π‘₯) 2 βˆ’ 92
π‘ˆπ‘ π‘–π‘›π‘” π‘Ž2 βˆ’ 𝑏2 = (π‘Ž + 𝑏)(π‘Ž βˆ’ 𝑏)
= (6π‘₯ + 9)(6π‘₯ βˆ’ 9)
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= 9(2π‘₯ + 3)(2π‘₯ βˆ’ 3)
b) πŸπŸ•π’™πŸ–π’šπŸ— + πŸ’πŸ“π’™πŸ—π’šπŸ“
Sol.
27π‘₯8𝑦9 + 45π‘₯9𝑦5
= 9π‘₯ 8 𝑦5(3𝑦4 + 5π‘₯)
c) π’™πŸ + πŸπ’™ βˆ’ πŸπŸ“
Sol.
π‘₯2 + 2π‘₯ βˆ’ 15
= π‘₯ 2 + 5π‘₯ βˆ’ 3π‘₯ βˆ’ 15
= (π‘₯ + 5)(π‘₯ βˆ’ 3)
5. Solve each inequality and write your answer in interval notation.
a) βˆ’πŸ < πŸ•π’™βˆ’πŸ“
πŸ‘ ≀ πŸ‘
Sol.
βˆ’2 < 7π‘₯ βˆ’ 5
3
βˆ’6 < 7π‘₯ βˆ’ 5
6 > 5 βˆ’ 7π‘₯
1 > βˆ’7π‘₯
βˆ’1 < 7π‘₯
π‘₯ > βˆ’1
7
7π‘₯ βˆ’ 5
3 ≀ 3
7π‘₯ βˆ’ 5 ≀ 9
7π‘₯ ≀ 14
π‘₯ ≀ 2
βˆ’1
7 < π‘₯ ≀ 2
b) |πŸ•π’™ + πŸ“| > 𝟐

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Sol.
|7π‘₯ + 5| > 2
βˆ’2 < 7π‘₯ + 5 < 2
7π‘₯ + 5 > βˆ’2
7π‘₯ > βˆ’7
π‘₯ > βˆ’1
7π‘₯ + 5 < 2
7π‘₯ < βˆ’3
π‘₯ < βˆ’3
7
βˆ’1 < π‘₯ < βˆ’3
7
c) πŸ‘π’™ + πŸ“ < πŸ–π’™ ≀ πŸπŸ“
Sol.
8π‘₯ > 3π‘₯ + 5
5π‘₯ > 5
π‘₯ > 1
8π‘₯ ≀ 25
π‘₯ ≀ 25
8
1 < π‘₯ ≀ 25
8
6. The area of the front of a book is πŸ”πŸŽ π’„π’ŽπŸ. The length of the book is πŸ• π’„π’Ž longer than
the width.
a) If the length of the book is 𝒙 π’„π’Ž, find the quadratic equation for the area of the
book 𝑨 using only the variables 𝒙 and 𝑨.
Sol.
𝐴 = π‘₯ βˆ— π‘€π‘–π‘‘π‘‘β„Ž … … … (1)
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π‘₯ = π‘€π‘–π‘‘π‘‘β„Ž + 7
π‘€π‘–π‘‘π‘‘β„Ž = 7 βˆ’ π‘₯ … … … (2)
πΉπ‘Ÿπ‘œπ‘š πΈπ‘ž(1) π‘Žπ‘›π‘‘ πΈπ‘ž(2)
𝐴 = π‘₯ (7 βˆ’ π‘₯)
𝐴 = 7π‘₯ βˆ’ π‘₯ 2
π‘₯2 βˆ’ 7π‘₯ + 𝐴 = 0
b) What are the dimensions of the book?
Sol.
𝐿𝑒𝑑 π‘‘β„Žπ‘’ π‘‘π‘–π‘šπ‘’π‘›π‘ π‘–π‘œπ‘›π‘  π‘œπ‘“ π‘‘β„Žπ‘’ π‘π‘œπ‘œπ‘˜ 𝑏𝑒 𝑙 π‘π‘š π‘Žπ‘›π‘‘ 𝑏 π‘π‘š π‘Ÿπ‘’π‘ π‘π‘’π‘π‘‘π‘–π‘£π‘’π‘™π‘¦
𝐴𝑐𝑐 π‘‘π‘œ π‘‘β„Žπ‘’ π‘žπ‘’π‘’π‘ π‘‘π‘–π‘œπ‘›,
𝑙𝑏 = 60 … … … (1)
𝑙 = 𝑏 + 7 … … … (2)
πΉπ‘Ÿπ‘œπ‘š πΈπ‘ž(1) π‘Žπ‘›π‘‘ πΈπ‘ž(2)
(𝑏 + 7)(𝑏) = 60
𝑏2 + 7𝑏 βˆ’ 60 = 0
𝑏2 + 12𝑏 βˆ’ 5𝑏 βˆ’ 60 = 0
𝑏(𝑏 + 12) βˆ’ 5(𝑏 + 12) = 0
(𝑏 + 12)(𝑏 βˆ’ 5) = 0
𝑏 = βˆ’12 π‘œπ‘Ÿ 𝑏 = 5
∡ π‘€π‘–π‘‘π‘‘β„Ž π‘π‘Žπ‘› π‘›π‘œπ‘‘ 𝑏𝑒 π‘›π‘’π‘”π‘Žπ‘‘π‘–π‘£π‘’, 𝑀𝑒 π‘–π‘”π‘›π‘œπ‘Ÿπ‘’ π‘›π‘’π‘”π‘Žπ‘‘π‘–π‘£π‘’ π‘£π‘Žπ‘™π‘’π‘’
𝑏 = 5
πΉπ‘Ÿπ‘œπ‘š πΈπ‘ž(2)
𝑙 = 5 + 7 = 12
π·π‘–π‘šπ‘’π‘›π‘ π‘–π‘œπ‘›π‘ : π‘™π‘’π‘›π‘”π‘‘β„Ž = 12π‘π‘š, π‘€π‘–π‘‘π‘‘β„Ž = 5π‘π‘š
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