Beginning Algebra, 11th Edition

Dividing Complex Numbers

Write each quotient in standard form.

(a)

Multiply numerator and denominator by the

conjugate of the denominator.

Multiply.

or - Combine like terms; ac-b=ca-bc

9

5

-

2

5

= i

• 9 - 2 i
  5

,

=

• 8 - 4 i+ 2 i- 1
  4 - 1 - 12

=

• 8 - 4 i+ 2 i+i^2
  4 - i^2

=

• 4 +i
  2 - i

#^2 +i

2 +i

• 4 +i
  2 - i

EXAMPLE 4

578 CHAPTER 9 Quadratic Equations

Solving a Quadratic Equation with Complex Solutions

(Quadratic Formula)

Solve for complex solutions.

Write the equation in standard form as

Quadratic formula

Substitute

and c=5.

a=2,b=-4,

x=

- 1 - 42  21 - 422 - 4122152

2122

x=

• b 2 b^2 - 4 ac
  2 a

2 x^2 - 4 x+ 5 =0.

2 x^2 = 4 x- 5

EXAMPLE 6

Extend the square root

property for k 60.

NOW TRY
EXERCISE 4
Write each quotient in stan-
dard form.

(a) (b)

5 +i

• i

3 - i

2 + 3 i

NOW TRY ANSWERS

4. (a) 133 - 1311 i (b)- 1 + 5 i

Be careful

with signs.

(b)

The conjugate of is or i.

Multiply.

; commutative property

=- 1 + 3 i

= i^2 =- 1

• 1 + 3 i
  
  
  • 1 - 12
  
  
  
  

=

3 i+i^2

• i^2

= 0 - i 0 +i,

3 +i

• i

#i

i

3 +i

• i

Be careful

with signs. NOW TRY

OBJECTIVE 5 Solve quadratic equations with complex number solutions.

Solving a Quadratic Equation with Complex Solutions

(Square Root Property)

Solve for complex solutions.

or

or

or Add

The solution set is 5 - 3  5 i 6. NOW TRY

x =- 3 + 5 i x=- 3 - 5 i -3.

x+ 3 = 5 i x + 3 = - 5 i 2 - 25 = 5 i

x + 3 = 2 - 25 x + 3 = - 2 - 25

1 x+ 322 =- 25

1 x+ 322 =- 25

NOW TRY EXAMPLE 5
EXERCISE 5
Solve for
complex solutions.

1 x- 122 =- 49

5. 51  7 i 6

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