Carnegie Learning Algebra II Student Skills Practice 1st Edition Chapter 2 Skills Practice
Page 303 Problem 1 Answer
Given: The number i
i=√−1
⇒ A number such that its square equals−1
Option E: A number such that its square equals−1
Page 303 Problem 2 Answer
Given: Imaginary number
Applying the above-given definition,
A number of the form bi Where b is not equal to 0
Option H: A number of the form bi where b is not equal to 0
Page 303 Problem 3 Answer
We need to match the definition of the pure imaginary number with the several available options.
Among the several available options, the correct option for the pure imaginary number is H.
That is, a pure imaginary number is a number of the form bi where b is not equal to 0.
Therefore, the correct option for the pure imaginary number is H.
Page 303 Problem 4 Answer
We need to match the definition of the complex numbers with the several available options.
Among the several available options, the correct option for the definition of a complex number is F.
That is, a complex number is a number in the form a+ib where a and b are real numbers.
Therefore, the correct option for the definition of a complex number is F.
Page 303 Problem 5 Answer
We need to match the definition of the real part of a complex number with the several available options.
Among the several available options, the correct option for the real part of a complex number is B.
That is, the real part of a complex number is the term a of a number written in the form a+bi.
Therefore, the correct option for the real part of a complex number is B.
Page 303 Problem 6 Answer
We need to match the definition of the immaginary part of a complex number with the several available options.
Among the several available options, the correct option for the immaginary part of a complex number is I.
That is, the immaginary part of a complex number is a term bi of a number written in the form a+bi.
Therefore, the correct option for the immaginary part of a complex number is I.
Page 303 Problem 7 Answer
We need to match the definition of the complex conjugate with the several available options.
Among the several available options, the correct option for the complex conjugate is D.
That is, the complex conjugate is a pair of numbers of the form a+bi and a−bi.
Therefore, the correct option for the complex conjugate is D.
Page 303 Problem 8 Answer
We need to match the definition of the monomial with the several available options.
Among the several available options, the correct option for the monomial is J.
That is, monomial is a polynomial with one term.
Therefore, the correct option for the monomial is J.
Page 303 Problem 9 Answer
We need to match the definition of the binomial with the several available options.
Among the several available options, the correct option for the binomial is C.
That is, binomial is a polynomial with two terms.
Therefore, the correct option for the binomial is C.
Page 303 Problem 10 Answer
We need to match the definition of the trinomial with the several available options.
Among the several available options, the correct option for the trinomial is G.
That is, a trinomial is a polynomial with three terms.
Therefore, the correct option for the trinomial is G.
Page 304 Problem 11 Answer
We need to find the value of i48.
We know that i2=−1.
So, i48=(i2)24
i48=(−1)24
i{48}=1
Therefore, the value of i48 is 1.
Page 304 Problem 12 Answer
Given imaginary number is i55.
To Do: Calculate the power.
i55=(i4)13⋅i3
=(1)13⋅(−i)
∵i4=1 and i3=−i
=(1)⋅(−i)
=−i
The power of i55=−i.
Page 303 Problem 13 Answer
Given imaginary number is i1000.
To Do: Calculate the power.
i1000=(i4)250
=(1)250
∵i4=1
=1
The power of i1000=1.
Page 304 Problem 14 Answer
Given imaginary number is i−22.
To Do: Calculate the power.
i−22=(i−2)11
=(−1)11
∵i−2=−1
=−1
The power of i−22=−1.
Page 304 Problem 15 Answer
Given imaginary number is i−7.
To Do: Calculate the power.
i−7=(i−2)3⋅(i)−1
=(−1)3⋅(−i)
∵i−2=−1 and i−1=−i
=(−1)⋅(−i)
=i
The power of i−7=i.
Page 305 Problem 16 Answer
Given number is √−72
To Do: Convert the number using imaginary number i.
√−72 =√36⋅2⋅(−1)
=6√2i
∵√36 =6 and √−1=i
√−72=6√2i
Page 305 Problem 17 Answer
Given number is 38−√−200+√121
To Do: Convert the number using imaginary number i.
38−√−200+√121
=38−√100⋅2⋅(−1)+√121
=38−10√2⋅(−1)+11
∵√100 =10 and √121
=11 and √−1
=i
=49−10√2/i
The required expression for38−√−200+√121
=49−10√2/i
Page 305 Problem 18 Answer
Given number is √−45+21
To Do: Convert the number using imaginary number i.
√−45 + 21 = √9 ⋅ 5 ⋅ (−1) + 21
= 3√5i + 21 ∵ √−1 = i and √9 = 3
The required expression for √−45 + 21 = 3√5i + 21
Page 305 Problem 19 Answer
Given number is √−48−12/4
To Do: Convert the number using imaginary number i.
√−48 − 12/4
√16 ⋅ 3 ⋅ (−1) − 12/4
=4√3i − 12/4
∵ √−1 = i and √16 = 4
=4(√3i − 3)/4
= √3i − 3
√−48 − 12/4 = √3i − 3
Page 305 Problem 20 Answer
Given number is 1+√4−√−15/3
To Do: Convert the number using imaginary number i.
=3−√15/i3
∵√−1=i
=1−√5/3i
The required expression for 1+√4−√−15/3
=1−√5/3i
Page 305 Problem 21 Answer
Given : −√−28+√21/3−√12/6
We have to rewrite the above expression using “i”.
−√−28 + − = − + − √21/3√12/6 √4 × 7 × (−1) √21/3√4 × 3/6
= −(√4 × √7 × √−1) + − √21/3√4 × √3/6
= −(2 × √7 × i) + − √21/3/2 × √3/2 × 3 [∵ i = √−1 ]
= −2√7i + − √21/3√3/3
= −2√7i +(√21 − √3)/3
The required expression for −√−28 + − = −2 i + √21/3√12/6 √7(√21 − √3)/3
Page 305 Problem 22 Answer
Given : √−75+√80/10
We have to rewrite the above expression using “i
“.√−75 + √80/10
√25 × 3 × (−1) + √16 × 5/10=
=(√25×√3×√−1)+(√16×√5)/10
=(5×√3×i)+(4×√5)/10 [∵ i=√−1 ]
=5×√3×i10+4×√5/10
=5×√3×I/5×2+2×2×√5/5×2
=√3×i2+2×√5/5
=√3i/2+2√5/5
The required expression for√−75+√80/10=√3i/2+2√5/5
Page 306 Problem 23 Answer
Given : (2+5i)−(7−9i)
We have to simplify the above expression.
(2+5i)−(7−9i)=2+5i−7+9i
=(2−7)+(5i+9i)
=−5+14i
The required expression for(2+5i)−(7−9i)=−5+14i
Page 306 Problem 24 Answer
Given : −6+8i−1−11i+13
We have to simplify the above expression.
−6 + 8i − 1 − 11i + 13 = (−6 − 1 + 13) + (8i − 11i)
=6+(−3i)
=6−3i
The required expression for −6+8i−1−11i+13=6−3i
Page 306 Problem 25 Answer
Given : −(4i−1+3i)+(6i−10+17)
We have to simplify the above expression.
−(4i − 1 + 3i) + (6i − 10 + 17) = −4i + 1 − 3i + 6i − 10 + 17
= (1 − 10 + 17) + (−4i − 3i + 6i)
= 8 + (−i)
= 8 − i
−(4i − 1 + 3i) + (6i − 10 + 17) = 8 − i
Page 306 Problem 26 Answer
Given : 22i+13−(7i+3+12i)+16i−25
We have to simplify the above expression.
22i+13−(7i+3+12i)+16i−25=22i+13−7i−3−12i+16i−25
=(13−3−25)+(22i−7i−12i+16i)
=−15+19i
22i+13−(7i+3+12i)+16i−25 =−15+19i
Page 306 Problem 27 Answer
Given : 9+3i(7−2i)
We have to simplify the above expression.
9+3i(7−2i)=9+(3i×7)−(3i×2i)
=9+21i−6i2
=9+21i−6(−1) [∵ i=√−1⇒i2=−1 ]
=9+21i+6
=(9+6)+21i
=15+21i
9+3i(7−2i)= =15+21i
Page 306 Problem 28 Answer
Given : (4−5i)(8+i)
We have to simplify the above expression.
(4−5i)(8+i)=4(8+i)−5i(8+i)
=(4×8+4×i)−(5i×8+5i×i)
=(32+4i)−(40i+5i2)
=32+4i−40i−5i2
=32+4i−40i−5(−1) [∵ i=√−1⇒i2=−1 ]
=32+4i−40i+5
=(32+5)+(4i−40i)
=37−36i
The required expression for(4−5i)(8+i)=37−36i
Page 306 Problem 29 Answer
Given : −0.5(14i−6)−4i(0.75−3i)
We have to simplify the above expression.
−0.5(14i−6)−4i(0.75−3i)=−(0.5×14i−0.5×6)−(4i×0.75−4i×3i)
=−(7i−3)−(3i−12i2)
=−7i+3−3i+12i2
=−7i+3−3i+12(−1) [∵ i=√−1⇒i2=−1 ]
=−7i+3−3i−12
=(3−12)+(−7i−3i)
=−9−10i
The required expression for−0.5(14i−6)−4i(0.75−3i)=−9−10i
Page 306 Problem 30 Answer
Given : (1/2i−3/4)(1/8−3/4i)
We have to simplify the above expression.
(1/2i−3/4)(1/8−3/4i)=1/2i(1/8−3/4i)−3/4
(1/8−3/4i) =(1/2i×1/8−1/2i×3/4i)−(3/4×1/8−3/4×3/4i)
=(1/16i−3/8i2)−(3/32−9/16i)
=1/16i−3/8i2−3/32+9/16i
=1/16i−3/8(−1)−3/32+9/16i [∵ i=√−1⇒i2=−1 ]
=1/16i+3/8−3/32+9/16i
=(3/8−3/32)+(1/16i+9/16i)
=(3×4−3/32)+(1/16+9/16)i
=(12−3/32)+(1+9/16)i
=9/32+10/16i
=9/32+5/8i
(1/2i−3/4)(1/8−3/4i)=9/32+5/8i
Page 307 Exercise 1 Answer
To find the product of(3+i)(3−i)
We have to multiply (3+i)(3−i)
First, open the brackets.
⇒(3+i)(3−i)
⇒3(3−i)+i(3−i)
⇒9−3i+3i−i2
⇒9−i2
⇒9−(−1)
⇒9+1
⇒10
The product of (3+i)(3−i)=10
Page 307 Exercise 2 Answer
To find the product of(4i−5)(4i+5)
We have to multiply (4i−5)(4i+5)
First, open the brackets.
⇒(4i−5)(4i+5)
⇒4i(4i+5)−5(4i+5)
⇒16i2+20i−20i−25
⇒16i2−25
⇒16(−1)−25
⇒−16−25
⇒−41
The product of (4i−5)(4i+5)=−41
Page 307 Exercise 3 Answer
To find the product of(1/3+3i)(1/3−3i)
We have to multiply (1/3+3i)(1/3−3i)
First, open the brackets.
⇒(1/3+3i)(1/3−3i)
⇒1/3(1/3−3i)+3i(1/3−3i)
⇒1/9−i+i−9i2
⇒1/9−9i2
⇒1/9−9(−1)
⇒1/9+9
⇒82/9
The product of (1/3+3i)(1/3−3i)=82/9
Page 307 Exercise 4 Answer
To find the product of(0.1+0.6i)(0.1−0.6i)
First, open the brackets.
⇒(0.1+0.6i)(0.1−0.6i)
⇒0.1(0.1−0.6i)+0.6i(0.1−0.6i)
⇒0.01−0.06i+0.06i−0.36i2
⇒0.01−0.36i2
⇒0.01−0.36(−1)
⇒0.01+0.36
⇒0.37
The product of (0.1+0.6i)(0.1−0.6i)=0.37
Page 307 Exercise 5 Answer
To identify the expression 4xi+7x as monomial, binomial or trinomial
The expression is 4xi+7x
This can be rewritten as ⇒(7+4i)x
There is only one x term. It is a monomial
The expression 4xi+7x is a monomial
Page 307 Exercise 6 Answer
To identify the expression−3x+5−8xi+1 as monomial, binomial or trinomial
The expression is −3x+5−8xi+1
⇒−3x+5−8xi+1
⇒6+(−3−8i)x
⇒6−(3+8i)x
There are two terms 6 and (3+8i)x. So this is a binomial.
The expression−3x+5−8xi+1 is a binomial
Page 307 Exercise 7 Answer
To identify the expression 6x2i+3x2 as monomial, binomial or trinomial
The expression is 6x2i+3x2
⇒6x2i+3x2
⇒(6i+3)x2
There is only one term. So this is a monomial.
The expression 6x2i+3x2 is a monomial
Page 307 Exercise 8 Answer
To identify the expression 8i−x3+7x2i as monomial, binomial or trinomial
The expression is 8i−x3+7x2i
⇒8i−x3+7x2i
⇒−x3+7ix2+8i
There are three terms x3,7ix2 and 8i. So this is a trinomial
The expression 8i−x3+7x2iis a trinomial
Page 308 Exercise 9 Answer
Given: The expression is xi−x+i+2−4i.
To identify that the expression is monomial, binomial, or trinomial.
The expression can be written as: xi−x+i+2−4i=x(i−1)+2−3i
Since there is only one term so the expression is monomial.
The expression xi−x+i+2−4i is monomial.
Page 308 Exercise 10 Answer
Given: The expression is−3x3i−x2+6x3+9i−1.
To identify that the expression is monomial, binomial, or trinomial.
The expression can be written as:−3x3i−x2+6x3+9i−1=3x3(2−i)−x2+9i−1
SInce here the highest term is 3 so the expression is trinomial.
The expression−3x3i−x2+6x3+9i−1 is monomial.
Page 308 Exercise 11 Answer
Given: The expression is (x−6i)2.
Simplify the expression.
Simplify the expression
(x−6i)2=x2+(6i)2−2x(6i)
(x−6i)2=x2+36i2−12xi
(x−6i)2=x2−12xi−36
The simplified expression is(x−6i)2
=x2−12xi−36.
Page 308 Exercise 12 Answer
Given: The expression is(2+5xi)(7−xi).
Simplify the expression.
Simplify the expression
(2+5xi)(7−xi)=2(7−xi)+5xi(7−xi)
(2+5xi)(7−xi)=14−2xi+35xi−5x2i2
(2+5xi)(7−xi)=14+33xi+5x2
The simplified expression is(2+5xi)(7−xi)=14+33xi+5x2.
Page 308 Exercise 13 Answer
Given: The expression is 3xi−4yi.
Simplify the expression.
Simplification of the expression by taking i common,
3xi−4yi=i(3x−4y)
The simplified expression is 3xi−4yi=i(3x−4y).
Page 308 Exercise 14 Answer
Given: The expression is(2xi−9)(3x+5i).
Simplify the expression.
Simplify the expression
(2xi−9)(3x+5i)=2xi(3x+5i)−9(3x+5i)
(2xi−9)(3x+5i)=6x2i+5xi2−27x−45i
(2xi−9)(3x+5i)=i(6x2−45)−32x
The simplified expression is(2xi−9)(3x+5i)=i(6x2−45)−32x.
Page 309 Exercise 15 Answer
Given: The expression is(x+4i)(x−4i)(x+4i).
Simplify the expression.
Simplify the expression (x+4i)(x−4i)(x+4i)=(x+4i)(x2−16i2)
⇒(x+4i)(x−4i)(x+4i)=(x+4i)(x2+16)
⇒(x+4i)(x−4i)(x+4i)=x3+16x+4x2i+64i
The simplified expression is(x+4i)(x−4i)(x+4i)=x3+16x+4x2i+64i.
Page 309 Exercise 16 Answer
Given: The expression is(3i−2xi)(3i−2xi)+(2i−3xi)(2i−3xi).
Simplify the expression.
Simplify the expression
(3i−2xi)(3i−2xi)+(2i−3xi)(2i−3xi)=(3i−2xi)2+(2i−3xi)2
(3i−2xi)(3i−2xi)+(2i−3xi)(2i−3xi)=(9i2+4x2i2−12xi2)+(4i2+9x2i2−12xi2)
(3i−2xi)(3i−2xi)+(2i−3xi)(2i−3xi)=(−9−4x2+12x)+(−4−9x2+12x)
(3i−2xi)(3i−2xi)+(2i−3xi)(2i−3xi)=24x−13x2−13
The simplified expression is(3i−2xi)(3i−2xi)+(2i−3xi)(2i−3xi)=24x−13x2−13.
Page 309 Exercise 17 Answer
Given: The complex number is 7+2i.
To find the conjugate of the complex number.
Find the conjugate of the complex number
The conjugate of a complex number can be found by changing the sign or the operator of the imaginary number.
So, the conjugate will be 7−2i.
The conjugate of the complex number is 7−2i.
Page 309 Exercise 18 Answer
Given: The complex number is 3+5i.
To find the conjugate of the complex number.
Find the conjugate of the complex number
The conjugate of a complex number can be found by changing the sign or the operator of the imaginary number.
So, the conjugate will be3−5i.
The conjugate of the complex number is 3−5i.
Page 309 Exercise 19 Answ
Given : The Complex Number z=8i
To Find: The conjugate of the complex number zˉ
Given the complex number z=8i———————— (1)
We know that z=a+ib
Here,
a= Real part
b = Imaginary part
From (1) we know that a=0 and b=8i
We also know that the conjugate of z is zˉ
=a−ib—————(2)
Hence to find the conjugate of the complex number sub a and b in (2)
Conjugate of the Complex Number zˉ
=a−ib
=0−i8
zˉ=−8i
We found that the conjugate of 8i is −8i
Page 309 Exercise 20 Answer
Given: The complex number z= −7i
To Find: The conjugate of the complex number zˉ
We know that z=a+ib
Here a= Real part and b= Imaginary part
From the given data z=−7i
We get a=0 and b=−7i
Hence to find the conjugate of the complex number zˉ
=a−ib—————–(1)
Sub the values of a and b in (1) zˉ
=a−ib
=0−(−7i)zˉ
=7i
We have found the conjugate of the complex number −7i is 7i
Page 309 Exercise 21 Answer
Given: The complex number z=2−11i
To Find: The conjugate of the complex number zˉ
We know z=a+ib
Here a= Real part and b= Imaginary part
Given the complex number z=2−11i
Here a=2 and b=−11i
To find the conjugate of the complex number zˉ
=a−ib——————-(1)
Sub the values of a and b in (1) zˉ
=a−ib zˉ
=2−(−11i)zˉ
=2+11i
We have found the conjugate of the complex number 2−11i is 2+11i
Page 309 Exercise 22 Answer
Given: The complex number z=−13−6i
To Find: The conjugate of the complex number zˉ
We know that z=a+ib
Here a= Real part and b= Imaginary part
From the given data z=−13−6i
We get a=−13 and b=−6i
To find the conjugate of the complex number zˉ
=a−ib——————(1)
Sub the values of a and b in (1)zˉ
=a−ibzˉ
=−13−(−6i)zˉ
=−13+6i
We have found the conjugate of the complex number −13−6i is −13+6i
Page 309 Exercise 23 Answer
Given: The complex number z=−21+4i
To Find: The conjugate of the complex number zˉ
We know that z=a+ib
Here a= Real part and b= Imaginary part
From the given data z=−21+4i
We get a=−21 and b=4i
To find the conjugate of the complex number zˉ
=a−ib—————-(1)
Sub the values of a and b in (1) zˉ
=a−ib zˉ
=−21−(4i)zˉ
=−21−4i
We have found the conjugate of the complex number −21+4i is −21−4i
Page 310 Exercise 24 Answer
Given: The complex number z=3+4i/5+6i
To Calculate: Each quotient
From the given data z=3+4i/5+6i
we can get
a=3
b=4
c=5
d=6
We know the formula a+ib/c+id={ac+bd/c2+d2}+i{bc−ad/c2+d2}
sub the values in this formula ={(3)(5)+(4)(6)/52+62}+i{(4)(5)−(3)(6)/52+62}
=15+24/61+i{20−18/61}
=39/61+i 2/61
Hence the solution for the given complex number is 39/61+i2/61
Page 310 Exercise 25 Answer
Given: The complex number z=8+7i/2+i
To Calculate: Each quotient
From the given data z=8+7i/2+i
we get
a=8
b=7
c=2
d=1
We know the formula a+ib/c+id={ac+bd/c2+d2}+i{bc−ad/c2+d2}
sub the above found values in the formula =(8)(2)+(7)(1)/22+12+i{(7)(2)−(8)(1)/22+12}
=16+7/5+i14−8/5
=23/5+i6/5
Hence the solution for the given complex number is 23/5+i6/5
Page 310 Exercise 26 Answer
Given: The complex number z=−1+5i/1−4i
To Calculate: Each quotient
From the given data z=−1+5i/1−4i
we get
a=−1
b=5
c=1
d=−4
We know the formula a+I b/c+id
=ac+bd/c2+d2+i{bc−ad/c2+d2}
sub the above found values in the formula =(−1)(1)+(5)(−4)/12+(−4)2+i{(5)(1)−(−1)(−4)/12+(−4)2}
=−1−20/17+i5−4/17
=−21/17+i{1/17}
Hence the solution for the given complex number is −21/17+i{1/17}
Page 307 Exercise 27 Answer
We are given: 6−3i/2−i
We are required to calculate the quotient.
We will multiply the numerator and denominator by the complex conjugate of the denominator and simplify the equation.
Given: 6−3i/2−i
The complex conjugate of the denominator is 2+i.
Multiplying and dividing this, we get
6−3i/2−i×2+I/2+i
=(6−3i)(2+i)/22−i2
=12−6i+6i−3i2/2−(−1)
=12−3(−1)/2+1
=12+3/3
=15/3
=5
The quotient of 6−3i/2−i is 5
Page 310 Exercise 28 Answer
We are given: 4−2i/−1+2i
We are required to calculate the quotient.
We will multiply the numerator and denominator by the complex conjugate of the denominator and simplify the equation.
Given: 4−2i/−1+2i
The complex conjugate of the denominator is: −1−2i.
Multiplying and dividing this, we get
4−2i/−1+2i×−1−2i/−1−2i
=(4−2i)(−1−2i)/(−1)2−(2i)2
=−4+2i−8i+4i2/1−2i2
=−4−6i−4/1+2
=−8−6i/3
The quotient of 4−2i/−1+2i is −8−6i/3.