Simplifying a Radical Review

Simplify each radical and leave the answer in exact form.1.

2.

3.

48

63

72

16 3 4 3

9 7 3 7

36 2 6 2

How many Real Solutions?

2 Complex Solutions

TwoReal

Solutions

OneReal

Solutions

How did you determine your answer?

Looking at the number of times it touches or crosses the x axis

Imaginary unit • Not all Quadratic Equations have real-number

solutions.

• To overcome this problem, mathematicians created an expanded system of numbers using the imaginary unit

• The imaginary number is use to write the square root of any negative number.

i

i

i

Definition• For any positive real number b,

2 2 1b b b i bi

• Modular 4 Pattern

1

1 1 2

( 1) 11 i i1 1 1

11 1 1 1 1 1

, 1, ,1 i ii2i3i4i

Example 1 Solve: x²+ 16 = 02 16x 2 16x

4x i 4 , 4x i i

16x

1 16x

No

Linear

Term!

Complex numbers

Expression that contains a real number and a pure imaginary number in the form (a + bi)

5 + 2i

5 is the real number 2i is the imaginary part.

Complex Number System Graphic Organizer

Rational #’s Irrational #a = 0Pure

Imaginary

a bi

0

Imaginary #

b

0a

Is every real number a complex number?

Yes

Is every imaginary number a complex number?

YesIs every Complex number a real number?

NO

Life is complex. It

has real and

imaginary

components. 

Discriminant The expression b²- 4ac is called the

Discriminant of the equation ax² + bx + c = 0

From the discriminant we can tell the nature and number of solutions for any given quadratic function.

2 11: 4 5Ex x x 20 4 5x x

1

4

5

a

b

c

2

2

4

( 4) 4(1)(5)

16 20

4

b ac

2 imaginary Solutions

Discriminant Graphic Organizer

Type One Type Two Type Three

Value of the Discriminant:

b2- 4ac >0 b2- 4ac = 0 b2- 4ac < 0

Number and Type of Solutions:

Two Real Solutions

One Real Solution

Two Imaginary

Solutions

Number of Intercepts:

Twox-intercept

Onex-intercept

No x-intercept

Graph of

Example:

2y ax bx c

Find the discriminant. Give the number and type of solutions of the equation.

2

2

2

8 17 0

8 16 0

8 15 0

x x

x x

x x

Ex 2:

Ex 3:

Ex 4:

Disc b² - 4ac= (-8)²- 4(1)(17)= -4

-4<0 so Two imaginary solutions

Disc (-8)²- 4(1)(16)= 0

0=0 so One real solutions

Disc (-8)²-4(1)(15) = 4

4>0 so Two real solutions

Quadratic Formula• Objective:

– To use the quadratic formula to find the solutions.

2 4

2

b b acx

a

• Let a, b, and c be real numbers such that a ≠ 0.

• Use the following formula to find the solutions of

the equation ax² + bx+ c = 0 (Standard Form).

Can you Sing it?

• X equals the opposite of b plus or minus the square root of b squared minus four AC all over 2 A.

Yes you Can!Pop Goes The

Weasel!

Parts of the Quadratic Formula

ax² + bx + c = 0

2 4

2x

b b ac

a

Quadratic Formula

2

2

4ab

a

bx

c

Discriminant

x-value of the VertexWhat kind of

solutions and

how many?

Method to find solutions of a quadratic

equation.

Example 5• Solve using the Quadratic

formula

2 3 2x x

3 17 3 17,

2 2x x

2 3 2 0x x

1, 3, 2a b c

23 3 4(1)( 2)

2(1)x

3 9 8

2

3 17

2

x

x

Example 6• Solve using the Quadratic formula

225 18 12 9x x x

3

5x

225 30 9 0x x

230 ( 30) 4(25)(9)

2(25)x

30 900 900

50x

30 0

5030

050

x

x

Standard Form

Identify the values of a, b and c

Plug Values into the Quadratic Formula

25, 30, 9a b c

Simplify under the radical

Simplify the formula

Write the Solution(s)

Example 7• Solve using the

Quadratic Formula2 4 5x x

2 4 5 0

1, 4, 5

x x

a b c

24 ( 4) 4(1)(5)

2(1)x

4 16 20

2x

4 16 20

2x

4 4

2x

4 2

2

ix

4 2 4 2

2 2 2 2

i ix and

2 , 2x i x i

imaginary

Practice 8• Solve using the Quadratic

formula2 6 15x x 26 6 4(1)(15)

2(1)x

6 36 60

2x

6 24

2x

imaginary

6 2 6

2

ix

6 2 6 6 2 6

2 2 2 2

i ix and x

3 6, 3 6x i x i

2 6 15 0x x