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