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Transcript
```Simplifying a Radical Review
Simplify each radical and leave the
1.
48  16  3  4 3
2.
63
 97  3 7
3.
72
 36  2  6 2
How many Real Solutions?
2
Complex
Solutions
Two
Real
Solutions
One
Real
Solutions
Looking at the number of times it touches or crosses the x axis
Imaginary unit
i
• Not all Quadratic Equations have real-number
solutions.
•To overcome this problem, mathematicians
created an expanded system of numbers using
the imaginary unit i
•The imaginary number i is use to write the
square root of any negative number.
Definition
• For any positive real number b,
b 2  b 2  1  b  i  bi
• Modular 4 Pattern
i  1
i , 1, i ,1
i  1  1  1
3
2
 1
 1 i  i
i  1  1  ( 1)
2
i  1  1  1  1  1 1  1
4
Example 1
Solve: x²+ 16 = 0
x  16
2
x  16
x   16
2
x   1  16
x  i  4
x  4i, 4i
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
Is every real number
a complex number?
Yes
a  bi
Is every Complex
number a real number?
NO
Rational #’s
Irrational #
Is every imaginary
number a complex
number?
Yes
b0
Imaginary #
a=0
Pure
Imaginary
Non-pure
a0
Imaginary
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
Ex 11:  x  4 x  5
2
0  x  4x  5
b 2  4ac
a 1
2

(

4)
 4(1)(5)
b  4
16  20
c5
4
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
One Real
Two Imaginary
Solutions
Solution
Solutions
Two
One
No
x-intercept
x-intercept
x-intercept
Number of
Intercepts:
Graph of
y  ax 2  bx  c
Example:
Find the discriminant.
Give the number and type of solutions of
the equation.
Ex 2:
x  8 x  17  0
2
Disc
b² - 4ac= (-8)²- 4(1)(17)= -4
-4<0 so Two imaginary solutions
Ex 3:
x  8 x  16  0
2
Disc (-8)²- 4(1)(16)= 0
0=0 so
One real solutions
Ex 4:
x  8 x  15  0
2
Disc
(-8)²-4(1)(15) = 4
4>0 so Two real solutions
• Objective:
– To use the quadratic formula to find the solutions.
• 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).
b  b  4ac
x
2a
2
Can you Sing it?
Yes you Can!
Pop Goes The Weasel!
• X equals the opposite of b plus or
minus the square root of b squared
minus four AC all over 2 A.
ax² + bx + c = 0
b  b  4ac
x
2a
2
Method to
find solutions
equation.
b  b  4ac
x
2a
x-value of the Vertex
2
Discriminant
What kind of
solutions
and
how many?
Example 5
x  3x  2
2
• Solve using the Quadratic formula
x  3x  2  0
2
a  1, b  3, c  2
3  32  4(1)(2)
x
2(1)
3  9  8
x
2
3  17
x
2
3  17
3  17
x
,x 
2
2
Example 6
25 x 2  18 x  12 x  9
• Solve using the Quadratic formula
25 x  30 x  9  0
2
Standard Form
Identify the values of a, b and c
a  25, b  30, c  9
Plug Values into the Quadratic Formula
30  (30)  4(25)(9)
x
2(25)
2
30  900  900
x
50
Simplify the formula
30  0
x
50
30
x
0
50
Write the Solution(s)
3
x
5
Example 7
• Solve using the
x  4x  5
2
x  4x  5  0
2
4  16  20
x
2
4  4
x
2
imaginary
a  1, b  4, c  5
4  2i
x
2
4  (4)2  4(1)(5)
x
2(1)
4 2i
4 2i
x   and 
2 2
2 2
4  16  20
x
2
x  2  i, x  2  i
Practice 8
• Solve using the Quadratic formula
x  6 x  15  x  6 x  15  0
2
2
6  2i 6
x
2
6  62  4(1)(15)
x
2(1)
6  36  60
x
2
6  24
x
2
6 2i 6
6 2i 6
x

and x 

2
2
2
2
x  3  i 6, x  3  i 6
imaginary
```