Post on 19-Jan-2016
transcript
Entry Task
Simplify
1)
2) 6
Solving Quadratic Equations by the
Quadratic Formula Learning Target: I can use the quadratic formula
to solve a quadratic equation.
THE QUADRATIC FORMULA
1. When you solve using completing the square on the general formula you get:
2. This is the quadratic formula!3. Just identify a, b, and c then substitute
into the formula.
2 4
2
b b acx
a
2 0ax bx c
WHY USE THE QUADRATIC FORMULA?
The quadratic formula allows you to
solve ANY quadratic equation, even
if you cannot factor it.
An important piece of the quadratic
formula is what’s under the radical:
b2 – 4ac
This piece is called the discriminant.
WHY IS THE DISCRIMINANT IMPORTANT?
The discriminant tells you the number and types of
answers
(roots) you will get. The discriminant can be +, –, or 0
which actually tells you a lot! Since the discriminant is
under a radical, think about what it means if you have a
positive or negative number or 0 under the radical.
WHAT THE DISCRIMINANT TELLS YOU!
Value of the Discriminant
Nature of the Solutions
Negative 2 imaginary solutions
Zero 1 Real Solution
Positive – perfect square
2 Reals- Rational
Positive – non-perfect square
2 Reals- Irrational
Example #1
22 7 11 0x x
Find the value of the discriminant and describe the nature of the roots (real,imaginary, rational, irrational) of each quadratic equation. Then solve the equation using the quadratic formula)
1.
a=2, b=7, c=-11
Discriminant = 2
2
4
(7) 4(2)( 11)
49
137
88
b ac
Discriminant =
Value of discriminant=137
Positive-NON perfect square
Nature of the Roots – 2 Reals - Irrational
Example #1- continued
22 7 11 0x x
2
2
4
2
7 7 4(2)( 11)
2(
2, 7, 11
7 137 2 Reals - Irrational
4
2)
a b
b ac
a
c
b
Solve using the Quadratic Formula
2 6 5x x Ex. 1 Solve.
2 5 6 0x x 5
6
1
c
b
a
2( 5) ( 5) 4(1)
(1)
( )
2
6x
5 25 24
2x
5 1
2x
5 1
2x
3,2x
2 2 8 0x x Ex. 2 Solve.
2
8
1
c
b
a
2( 2) ( 2 (1)
(1)
() 4 )
2
8x
4, 2x
2 14 49 0x x Ex. 3 Solve.
14
4
1
9
b
a
c
2 (1)
(1
(14) (14) ( )
)
4
2
49x
7x
218 10 0x x Ex. 4 Solve.
10
18
1b
a
c
2 (14 8)
(
(1) (1)
82 1 )
(10)x
no solution
Solve x 2 + 9x + 14 = 0.
Use the quadratic formula.
1x 2 + 9x + 14 = 0
– b b 2 – 4( a )( c )2( a )
x =
–9 81 – 562
x =
–9 252
x =
Identify a = 1, b = 9, and c = 14.
Substitute values in the
quadratic formula.
Simplify.
The equation has two solutions:
–9 52
x =
9 9 1 14 1
Simplify.
–9 + 52
x =
–9 – 52
x =
= –2
= –7
SOLUTION
Using the Quadratic Formula
Check the solutions to x 2 + 9x + 14 = 0 in the original equation.
Check x = –2 : Check x = –7 :
x 2 + 9x + 14 = 0
0 = 0
(–2) 2 + 9(–2) + 14 = 0?
4 + –18 + 14 = 0?
x 2 + 9x + 14 = 0
0 = 0
(–7) 2 + 9(–7) + 14 = 0?
49 + –63 + 14 = 0?
Using the Quadratic Formula
Find the x-intercepts of the graph of y = –x 2 – 2 x + 5.
Finding the x-Intercepts of a Graph
The x-intercepts occur when y = 0.
0 = –1x 2 – 2 x + 5
–( b ) ( b ) 2 – 4( a )( c )2( a )
x =
2 4 + 20–2
x =
2 24–2
x =
Substitute 0 for y, and
identify a = 1, b = –2, and c = 5.
Substitute values in the
quadratic formula.
Simplify.
Solutions
The equation has two solutions:
2 24–2
x =
2 + 24–2
x = –3.45
2 – 24–2
x = 1.45
–2 –2 –1 5 –1
SOLUTION
y = –x 2 – 2 x + 5 Write original equation.
Check your solutions to the equation y = –x 2 – 2 x + 5 graphically.
Finding the x-Intercepts of a Graph
You can see that the graph shows the x-intercepts between –3 and –4 and between 1 and 2.
Check y –3.45 and y 1.45.
Using Quadratic Models in Real Life
Problems involving models for the dropping or throwing of an object are called
vertical motion problems.
VERTICAL MOTION MODELS
OBJECT IS DROPPED: h = –16 t 2 + s
h = –16 t 2 + v t + sOBJECT IS THROWN:
h = height (feet) t = time in motion (seconds)
s = initial height (feet) v = initial velocity (feet per second)
In these models the coefficient of t 2 is one half the acceleration due to gravity. On the surface of Earth, this acceleration is approximately 32 feet per second per second.
Remember that velocity v can be positive (object moving up), negative (object moving down), or zero (object not moving). Speed is the absolute value of velocity.
BALLOON COMPETITION
Because the marker is thrown down, the initial velocity is v = –30 feet per second. The initial height is s = 200 feet. The marker will hit the target when the height is 0.
v = –30, s = 200, h = 0.
SOLUTION
Modeling Vertical Motion
You are competing in the Field Target Event at a hot-air balloon festival. You throw a marker down from an altitude of 200 feet toward a target. When the marker leaves your hand, its speed is 30 feet per second. How long will it take the marker to hit the target?
BALLOON COMPETITION
v = –30, s = 200, h = 0.
h = –16 t 2 + v t + s
h = –16 t 2 + v t + s
h = –16 t 2 – 30t + 200
t 2.72 or –4.60
–( b ) ( b ) 2 – 4( a )( c )2( a )
t =
30 13,700–32
t =
–30 –30 –16 200 –16
Substitute values in the quadratic formula.
Substitute 0 for h. Write in standard form.
Substitute values for v and s into the
vertical motion model.
Choose the vertical motion model for a thrown object.
Simplify.
Solutions
(–30) 200
0
SOLUTION
Modeling Vertical Motion
BALLOON COMPETITION
t 2.72 or –4.60
As a solution, –4.60 does not make sense in the context of the problem.
SOLUTION
Modeling Vertical Motion
Therefore, the weighted marker will hit the target about 2.72 seconds after it was thrown.
Solving Quadratic Equations by the Quadratic Formula
2
2
2
2
2
1. 2 63 0
2. 8 84 0
3. 5 24 0
4. 7 13 0
5. 3 5 6 0
x x
x x
x x
x x
x x
Try the following examples. Do your work on your paper and then check your answers.
1. 9,7
2.(6, 14)
3. 3,8
7 34.
2
5 475.
6
i
i
Skip #4 and #5
Homework
Homework – P. 245
#12,15,18,21,23,24,25,28,37,38