1) 3

2) {x = 9 or x = -1}

3) (b) The modulus of the complex number 5-3i is

Step-by-step explanation:

Problem 1

3 - 2

i) Cancelling out the 2's from the top and bottom of the middle term, we get

3 -

ii) Cancelling out - and +, we get

3 as the simplified form

Problem 2

x-1 =

Our first goal is to get rid off the radical on the right side

i) Squaring both sides, we get

=

ii) (x-1)*(x-1) = 6x+10

iii) Applying the distributive property (a+b)(c+d) = ac+ad+bc+bd to the left side of the equation, we get

(x)(x)+(x)(-1)+(-1)(x)+(-1)(-1) = 6x+10

=> -x-x+1 = 6x+10

=> -2x+1 = 6x+10

iv) Subtract 6x from both sides, we get

-2x+1-6x = 6x+10-6x

v) Cancelling out 6x and -6x from the right side, we get

-2x-6x+1 = 10

=> -8x+1 = 10

vi) Subtracting 10 from both the sides, we get

-8x+1-10 = 10-10

vii) Cancelling out 10 and 10 from the right side, we have

-8x-10+1 = 0

=> 1-8x-9 = 0

viii) Coefficient of the first term = 1

Multiplying the coefficient of the first term and the last term, we get

1*(-9) = -9

We need to find out two such factors of -9 which when added should give the middle term -8

So, -9 and +1 are the two factors of -9 which when added gives us the middle term -8

ix) Rewriting the middle term, we get

-9x+x-9 = 0

x) Factoring out x from the first two terms and factoring out 1 from the last two terms, we get

x(x-9)+1(x-9)=0

xi) Factoring out x-9 from both the terms, we get

(x-9)(x+1)=0

xii) Either x-9=0 or x+1=0

xiii) Solving x-9=0, we get x=9

xiv) Solving x+1=0, we get x= -1

So, solution set {x = 9 or x = -1}

Problem 3

5 − 3i

a) In order to graph the complex number 5-3i, we need to move right by 5 units on the real axis and then move down by 3 units on the imaginary axis.

See figure attached

b) A complex number is in the form of z= a+ bi

i) Comparing 5-3i with a+bi, we get a=5 and b = -3

The modulus is given by

|z| =

ii) Plugging in a=5 and b=-3, we get

|z| =

iii) |z| =

iv) |z| =

The modulus of the complex number 5-3i is

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