Expand and simplify expressions
VerifiedAdded on Β 2023/04/20
|6
|1031
|264
AI Summary
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.
Contribute Materials
Your contribution can guide someoneβs learning journey. Share your
documents today.
DIP1003 SEMESTER 1,2019
ASSIGNMENT 2
MARCH 26, 2019
Submitted by:
Timo Weiss
ASSIGNMENT 2
MARCH 26, 2019
Submitted by:
Timo Weiss
Secure Best Marks with AI Grader
Need help grading? Try our AI Grader for instant feedback on your assignments.
1. Expand and simplify the following expressions:
a) (ππ + π)π β ππ(π + ππ)
Sol.
Using (π + π)2 = π2 + π2 + 2ππ
(4π₯ + π¦)2 β 3π₯(π¦ + 2π₯)
= ((4π₯)2 + π¦2 + 2(4π₯)(π¦)) β 3π₯(π¦ + 2π₯)
= 16π₯ 2 + π¦2 + 8π₯π¦ β 3π₯π¦ β 6π₯ 2
= 10π₯ 2 + π¦2 + 5π₯π¦
b)
(ππππππ)
π
π(π
βπ
π π
π
π)
βπ
ππππππ
πβππππ
Sol.
(π₯4π¦0π§6)1
2(π₯β3
2 π¦6
2)β2π₯4π¦2π§3
π₯β4π¦π§3
= (π₯2π§3)π₯4π¦2π§3
(π₯β3π¦6)(π₯β4π¦π§3)
= π₯6π¦2π§6
π₯β7π¦7π§3
= π₯13 π¦β5π§3
= π₯13 π§3
π¦5
2. Fid the values of x in the following equation using the rule that if π β π = π then π =
π ππ π = π . Do not expand the equation or use the quadratic formula.
(ππ + π) (ππ β π
π) = π
Sol.
πΈππ‘βππ (5π₯ + 6) = 0 ππ (3π₯ β 2
3) = 0
5π₯ + 6 = 0
π₯ = β6
5
(3π₯ β 2
3) = 0
a) (ππ + π)π β ππ(π + ππ)
Sol.
Using (π + π)2 = π2 + π2 + 2ππ
(4π₯ + π¦)2 β 3π₯(π¦ + 2π₯)
= ((4π₯)2 + π¦2 + 2(4π₯)(π¦)) β 3π₯(π¦ + 2π₯)
= 16π₯ 2 + π¦2 + 8π₯π¦ β 3π₯π¦ β 6π₯ 2
= 10π₯ 2 + π¦2 + 5π₯π¦
b)
(ππππππ)
π
π(π
βπ
π π
π
π)
βπ
ππππππ
πβππππ
Sol.
(π₯4π¦0π§6)1
2(π₯β3
2 π¦6
2)β2π₯4π¦2π§3
π₯β4π¦π§3
= (π₯2π§3)π₯4π¦2π§3
(π₯β3π¦6)(π₯β4π¦π§3)
= π₯6π¦2π§6
π₯β7π¦7π§3
= π₯13 π¦β5π§3
= π₯13 π§3
π¦5
2. Fid the values of x in the following equation using the rule that if π β π = π then π =
π ππ π = π . Do not expand the equation or use the quadratic formula.
(ππ + π) (ππ β π
π) = π
Sol.
πΈππ‘βππ (5π₯ + 6) = 0 ππ (3π₯ β 2
3) = 0
5π₯ + 6 = 0
π₯ = β6
5
(3π₯ β 2
3) = 0
3π₯ = 2
3
π₯ = 2
9
π₯ = β6
5 ππ 2
9
3. Solve the following equation for π.
Hint: ππ β ππ = (π β π)(π + π)
(ππ)π β π
ππ + π = ππ + π
π + π
Sol.
(2π)2 β 4
π2 + 1 = 2π + 2
π + 2
= (2π)2 β 22
π2 + 1 = 2π + 2
π + 2
= (2π + 2)(2π β 2)
π2 + 1 = 2π + 2
π + 2
= 4(π + 1)(π β 1)
π2 + 1 = 2(π + 1)
π + 2
= 2(π β 1)
π2 + 1 = 1
π + 2
2(π β 1)(π + 2) = π2 + 1
2(π2 + 2π β π β 2) = π2 + 1
2π2 + π β 2 = π 2 + 1
= π2 + π β 3
4. Factorise the following expressions:
a) ππππ β ππ
Sol.
36π₯2 β 81
= (6π₯) 2 β 92
ππ πππ π2 β π2 = (π + π)(π β π)
= (6π₯ + 9)(6π₯ β 9)
3
π₯ = 2
9
π₯ = β6
5 ππ 2
9
3. Solve the following equation for π.
Hint: ππ β ππ = (π β π)(π + π)
(ππ)π β π
ππ + π = ππ + π
π + π
Sol.
(2π)2 β 4
π2 + 1 = 2π + 2
π + 2
= (2π)2 β 22
π2 + 1 = 2π + 2
π + 2
= (2π + 2)(2π β 2)
π2 + 1 = 2π + 2
π + 2
= 4(π + 1)(π β 1)
π2 + 1 = 2(π + 1)
π + 2
= 2(π β 1)
π2 + 1 = 1
π + 2
2(π β 1)(π + 2) = π2 + 1
2(π2 + 2π β π β 2) = π2 + 1
2π2 + π β 2 = π 2 + 1
= π2 + π β 3
4. Factorise the following expressions:
a) ππππ β ππ
Sol.
36π₯2 β 81
= (6π₯) 2 β 92
ππ πππ π2 β π2 = (π + π)(π β π)
= (6π₯ + 9)(6π₯ β 9)
= 9(2π₯ + 3)(2π₯ β 3)
b) ππππππ + ππππππ
Sol.
27π₯8π¦9 + 45π₯9π¦5
= 9π₯ 8 π¦5(3π¦4 + 5π₯)
c) ππ + ππ β ππ
Sol.
π₯2 + 2π₯ β 15
= π₯ 2 + 5π₯ β 3π₯ β 15
= (π₯ + 5)(π₯ β 3)
5. Solve each inequality and write your answer in interval notation.
a) βπ < ππβπ
π β€ π
Sol.
β2 < 7π₯ β 5
3
β6 < 7π₯ β 5
6 > 5 β 7π₯
1 > β7π₯
β1 < 7π₯
π₯ > β1
7
7π₯ β 5
3 β€ 3
7π₯ β 5 β€ 9
7π₯ β€ 14
π₯ β€ 2
β1
7 < π₯ β€ 2
b) |ππ + π| > π
b) ππππππ + ππππππ
Sol.
27π₯8π¦9 + 45π₯9π¦5
= 9π₯ 8 π¦5(3π¦4 + 5π₯)
c) ππ + ππ β ππ
Sol.
π₯2 + 2π₯ β 15
= π₯ 2 + 5π₯ β 3π₯ β 15
= (π₯ + 5)(π₯ β 3)
5. Solve each inequality and write your answer in interval notation.
a) βπ < ππβπ
π β€ π
Sol.
β2 < 7π₯ β 5
3
β6 < 7π₯ β 5
6 > 5 β 7π₯
1 > β7π₯
β1 < 7π₯
π₯ > β1
7
7π₯ β 5
3 β€ 3
7π₯ β 5 β€ 9
7π₯ β€ 14
π₯ β€ 2
β1
7 < π₯ β€ 2
b) |ππ + π| > π
Secure Best Marks with AI Grader
Need help grading? Try our AI Grader for instant feedback on your assignments.
Sol.
|7π₯ + 5| > 2
β2 < 7π₯ + 5 < 2
7π₯ + 5 > β2
7π₯ > β7
π₯ > β1
7π₯ + 5 < 2
7π₯ < β3
π₯ < β3
7
β1 < π₯ < β3
7
c) ππ + π < ππ β€ ππ
Sol.
8π₯ > 3π₯ + 5
5π₯ > 5
π₯ > 1
8π₯ β€ 25
π₯ β€ 25
8
1 < π₯ β€ 25
8
6. The area of the front of a book is ππ πππ. The length of the book is π ππ longer than
the width.
a) If the length of the book is π ππ, find the quadratic equation for the area of the
book π¨ using only the variables π and π¨.
Sol.
π΄ = π₯ β π€πππ‘β β¦ β¦ β¦ (1)
|7π₯ + 5| > 2
β2 < 7π₯ + 5 < 2
7π₯ + 5 > β2
7π₯ > β7
π₯ > β1
7π₯ + 5 < 2
7π₯ < β3
π₯ < β3
7
β1 < π₯ < β3
7
c) ππ + π < ππ β€ ππ
Sol.
8π₯ > 3π₯ + 5
5π₯ > 5
π₯ > 1
8π₯ β€ 25
π₯ β€ 25
8
1 < π₯ β€ 25
8
6. The area of the front of a book is ππ πππ. The length of the book is π ππ longer than
the width.
a) If the length of the book is π ππ, find the quadratic equation for the area of the
book π¨ using only the variables π and π¨.
Sol.
π΄ = π₯ β π€πππ‘β β¦ β¦ β¦ (1)
π₯ = π€πππ‘β + 7
π€πππ‘β = 7 β π₯ β¦ β¦ β¦ (2)
πΉπππ πΈπ(1) πππ πΈπ(2)
π΄ = π₯ (7 β π₯)
π΄ = 7π₯ β π₯ 2
π₯2 β 7π₯ + π΄ = 0
b) What are the dimensions of the book?
Sol.
πΏππ‘ π‘βπ ππππππ ππππ ππ π‘βπ ππππ ππ π ππ πππ π ππ πππ ππππ‘ππ£πππ¦
π΄ππ π‘π π‘βπ ππ’ππ π‘πππ,
ππ = 60 β¦ β¦ β¦ (1)
π = π + 7 β¦ β¦ β¦ (2)
πΉπππ πΈπ(1) πππ πΈπ(2)
(π + 7)(π) = 60
π2 + 7π β 60 = 0
π2 + 12π β 5π β 60 = 0
π(π + 12) β 5(π + 12) = 0
(π + 12)(π β 5) = 0
π = β12 ππ π = 5
β΅ π€πππ‘β πππ πππ‘ ππ πππππ‘ππ£π, π€π ππππππ πππππ‘ππ£π π£πππ’π
π = 5
πΉπππ πΈπ(2)
π = 5 + 7 = 12
π·πππππ ππππ : πππππ‘β = 12ππ, π€πππ‘β = 5ππ
π€πππ‘β = 7 β π₯ β¦ β¦ β¦ (2)
πΉπππ πΈπ(1) πππ πΈπ(2)
π΄ = π₯ (7 β π₯)
π΄ = 7π₯ β π₯ 2
π₯2 β 7π₯ + π΄ = 0
b) What are the dimensions of the book?
Sol.
πΏππ‘ π‘βπ ππππππ ππππ ππ π‘βπ ππππ ππ π ππ πππ π ππ πππ ππππ‘ππ£πππ¦
π΄ππ π‘π π‘βπ ππ’ππ π‘πππ,
ππ = 60 β¦ β¦ β¦ (1)
π = π + 7 β¦ β¦ β¦ (2)
πΉπππ πΈπ(1) πππ πΈπ(2)
(π + 7)(π) = 60
π2 + 7π β 60 = 0
π2 + 12π β 5π β 60 = 0
π(π + 12) β 5(π + 12) = 0
(π + 12)(π β 5) = 0
π = β12 ππ π = 5
β΅ π€πππ‘β πππ πππ‘ ππ πππππ‘ππ£π, π€π ππππππ πππππ‘ππ£π π£πππ’π
π = 5
πΉπππ πΈπ(2)
π = 5 + 7 = 12
π·πππππ ππππ : πππππ‘β = 12ππ, π€πππ‘β = 5ππ
1 out of 6
Related Documents
Your All-in-One AI-Powered Toolkit for Academic Success.
Β +13062052269
info@desklib.com
Available 24*7 on WhatsApp / Email
Unlock your academic potential
Β© 2024 Β | Β Zucol Services PVT LTD Β | Β All rights reserved.