Math 1314 Chapter 1 Homework Assignment: Detailed Solutions Guide

This document provides a comprehensive solution set for a Math 1314 Chapter 1 homework assignment. The solutions cover a wide range of algebra topics, including simplifying expressions involving complex numbers, performing operations with complex numbers, solving quadratic equations using various methods (zero-factor property, square root property, completing the square, and the quadratic formula), and solving equations for indicated variables. Furthermore, the solutions encompass absolute value equations, and provide detailed explanations for each step. The assignment also includes problems from Math 1314 HW 4 (Chapter 4) covering topics like functions, inverse functions, logarithmic and exponential forms, and the use of logarithmic properties.

1
Math 1314 Chapter 1 Homework
Question 1 Part a
− √−18=− √−1 ×2 × 9=−3 √2 √−1=−3 √ 2i
Question 1 Part b
√−3 √−8= √(−3)(−8)= √24= √ 4 × 6=2 √ 6
Question 1 Part c
√−30
√−10 = √ −30
−10 = √ 3
Question 1 Part d
√−6 √−2
√3 = √ (−6 ) (−2 )
3 = √ 12
3 = √4=2
Question 2
−3+ √−18
24 =−3+ √−1× 2× 9
24 =−3+3 i √2
24 =−1
8 + √2
8 i
Question 3-part a
( 3+2 i )+ ( 9−3 i )= (3+ 9 ) + ( 2 i−3 i )=12−i
Question 3-part b
( 2−5 i )− (3+4 i )− (−2+ i )= ( 2−3+2 )+ (−5i−4 i−i ) =1−10 i
Question 4 part a
( 3−2i ) 2= ( 3−2 i ) ( 3−2i ) =3 ( 3−2i ) −2i ( 3−2 i ) =9−6 i−6 i−4=5−12 i

2
Question 4 part b
( 2+4 i ) (−1+3 i )=2 (−1+3 i )+ 4 i (−1+ 3i ) =−2+6 i−4 i−12=−14 +2i
Question 5 part a
2−i
2+i =2−i
2+ i × 2−i
2−i = ( 2−i ) ( 2−i )
( 2+i ) ( 2−i ) =2 ( 2−i )−i ( 2−i )
2 ( 2−i ) +i ( 2−i )
¿ 4−2 i−2i−1
4−2 i+2i+1 =3−4 i
5 = 3
5 − 4 i
5 =0.6−0.8 i
Question 5 part b
2
3i = 2 ×3 i
3 i×3 i = 6i
−9 =−2
3 i
Question 6 part a
i22=(i¿¿ 2)11= ( −1 ) 11=−1¿
Question 6 part b
i32=(i ¿¿ 2)16= ( −1 ) 16=1 ¿
Question 7 part a
x2−5 x+ 6=0
x2−5 x+6=x2−2 x+ 3 x +6=0
x2−2 x−3 x+6=x ( x −2 )−3 ( x−2 )=0
( x−3 ) ( x−2 ) =0
( x−3 )=0 , x=3

3
¿ ( x−2 ) =0. x=2
x=2∨x=3
Question 7 part b
4 x2 −4 x +1=0
4 x2 −4 x +1=4 x2−2 x−2 x +1=2 x ( 2 x−1 ) −1 ( 2 x−1 ) =0
( 2 x−1 ) (2 x−1 )=0
2 x−1=0 , 2 x=1 , x= 1
2
Hence , x=1
2
Question 8 part a
x2=16 , x= √ x2 = √ 16
x=± 4
Question 8 part b
( 3 x−1 )2=9 x2−1=12
9 x2=1+12=13
x2= 13
9
√ x2=x= √ 13
9 =± √ 13
3

4
Question 9 part a
2 x2 −x−28=0
x2− x
2 −14=0
x2− x
2 =14
Take half of the coefficient of x and square it then add the result to both sides of the equation to
get:
( −1
2 . 1
2 )
2
= 1
16
x2− x
2 + 1
16 =14+ 1
16 = 225
16
(x− 1
4 )2
= 225
16
x=−1
4 ± 15
4
x=−1
4 + 15
4 =4∨x=−1
4 − 15
4 =−3.5
Question 9 part b
x2−2 x−2=0
x2−2 x=2
(−2 × 1
2 )
2
=1

5
x2−2 x+1=2+1
( x− 2
2 )
2
=3
( x−1 ) 2=3
x=1 ± √3=2.732∨−0.732
Question 10
x2=2 x−5
x2−2 x+5=0 , a=1 ,b=−2 , c=5
x=−b ± √ b2−4 ac
2 a =2 ± √(−2)2−4 (1 ×5)
2(1) =2 ± √−16
2 = 2 ± 4 i
2 =1± 2 i
x=1+2 i∨x=1−2 i
Question 11 part a
F × r
kM = kM v2
r × r
kM
v2= Fr
kM
V =± √ Fr
kM
Question 11 part b
2 x2 +4 xy−3 y2=2
2 x2 + 4 yx− ( 2+3 y2 ) =0

6
a=2 , b=4 y , c=− ( 2+3 y2 )
x=−b ± √ b2−4 ac
2 a =−4 y ± √ (−4 y)2 +4 ×2 ( 2+3 y2 )
2(2) =−4 y ± √ 16 y2 +16+24 y2
4
x=−4 y ± √ 40 y2 +16
4
Question 12
Let the integers be n ,n+ 2
n ( n+2 ) =224
n2 +2 n−224=0
n=−2± √ (−2)2−4(−224 )
2 =−2 ± 30
2 =−1± 15=14∨−16
The integers are14∧16∨−16∧−14
Question 13
Let the integers be n ,n+ 2
n2 + ( n+2 )2=202
n2 +n2+ 4 n+ 4=202
n2 +n2+ 4 n+4−202=0
2 n2 +4 n−198=0
2n2
2 + 4 n
2 − 198
2 = 0
2 =0

7
n2 +2 n−99=0
n=−2± √(−2)2−4(−99)
2 =−2 ±20
2
n=9∨n=−11
The numbers are 9∧11∧−11∧−9
Question 14
L=2 W
Perimeter=2 L+2 W
Area=LW
2 ( 2 L+2 W )=LW
4 L+ 4 W =LW
But L=2 W
4 (2 W )+4 W =(2W ) W
12 W=2 W 2
6 W =W 2
W 2
W =6 W
W
W =6
L=2 W =2× 6=12

8
Question 15
Let the length of the longer leg=n , shorter leg=90 m, hypoteneuse=30+n
( 30+n ) 2=n2 +902
900+60 n+ n2=n2+8100
60 n=8100−900=7200
n=7200
60 =120 m
Question 16
2 x +5
2 − 3 x
x−2 =x
( 2 x +5 ) ( x−2 )−3 x (2)
2 ( x−2 ) =x
2 x2 +5 x−4 x−10−6 x
2 x −4 =x
2 x2 +5 x−4 x−10−6 x=x(2 x−4)
2 x2 −5 x −10=2 x2−4 x
Rearranging the equation yields:
−5 x+ 4 x=10 , x=−10
Question 17
2 x +1
x−2 − 3
x = −6
x2−2 x

9
( 2 x +1 ) x−3( x−2)
( x −2) x = −6
x2−2 x
2 x2−2 x+ 6
x2−2 x = −6
x2 −2 x
2 x2 −2 x +6=−6
2 x2 −2 x +12=0
x= 2± √(2)2−4(2 ×12)
2(2) =−2± √−92
2 =−2± 2 √23 i
2 =−1 ± √23 i
x=−1+ √ 23 i∨x=−1− √ 23 i
Question 18
x− √2 x+3=0
x2=2 x+3
x2−2 x−3=0
x= 2± √ (2)2−4(1 ×−3)
2 = 2± 4
2 =1± 2
x=1+2=3∨x=1−2=−1
x=3∨x=−1
Question 19
√2 x +5− √x +2=1
√2 x +5=1+ √x +2

10
Squaring both sides:
( √ 2 x +5 ) 2= ( 1+ √ x +2 ) 2
2 x+5=1+2 √ x +2+x +2
2 x+5−1−x−2=1+2 √ x +2+ x +2−1−x −2
x +2=2 √ x+2
Square both sides to get:
( x +2 )2= ( 2 √x +2 )2
x2+ 4 x +4=4 ( x +2 )=4 x+8
x2+ 4 x +4−4 x−8=0
x2−4=0
x2=4
x=± 2
x=2∨x=−2
Question 20
( x−3 )
2
5 =4
( ( x−3 )
2
5 )5
2
=4
5
2
x−3= ( 22 )
5
2 =25=32

11
x=32+3=35
Question 21
2 x4−7 x2 +5=0
Let x2=u∧x4=u2
2 x4−7 x2 +5=2 u2−7 u+ 5=0
u=7 ± √ (7)2−4( 2×5)
2(2) = 7 ±3
4
u=7−3
4 =1∨u= 7+3
4 =2.5
x2=1∨x2=2.5
x=± 1∨x=± √2.5
x=1 , x=−1 , x= √2.5 , x=− √2.5
Question 22 part a
3 ( x+5 ) +1≥ 5+3 x
3 x+ 15+1≥ 5+3 x
3 x+ 16 ≥5+3 x
3 x−3 x ≥ 5−16
0 ≥−11
Hence, the solution is true for all x. That is (−∞, ∞) in interval notation.

12
Question 22 part b
2>−6 x +3>−3
2>−6 x +3∧−6 x+3>−3
2−3 >−6 x+3−3
−1>−6 x
x > −1
−6
x > 1
6
−6 x +3−3>−3−3
−6 x >−6
x ← 6
−6
x <1
Combining the two solutions we get:
1
6 < x<1
¿ interval notation ( 1
6 , 1)
Question 22 part c
4 x +7
−3 ≤2 x+5