1. Entry Task
Simplify
1) √80 2 2)

2. Answers

3. Solving Quadratic Equations by the Quadratic Formula
Learning Target:
I can use the quadratic formula to solve a quadratic equation.

4. Complete the Square and solve for x

5. THE QUADRATIC FORMULA When you solve by completing the square you get:
This is the quadratic formula!
Just identify a, b, and c then substitute into the formula.

6. 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.

7. 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.

8. 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

9. Example #1 a=2, b=7, c=-11 Discriminant = Discriminant =
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 =
Value of discriminant=137
Positive-NON perfect square
Nature of the Roots – 2 Reals - Irrational
Discriminant =

10. Example #1- continued
Solve using the Quadratic Formula

11. Solve.
Ex. 1

12. Solve.
Ex. 2

13. Solve.
Ex. 3

14. Solve.
Ex. 4

15. Using the Quadratic Formula
Solve x 2 + 9x + 14 = 0.
SOLUTION
Use the quadratic formula.
1x 2 + 9x + 14 = 0
Identify a = 1, b = 9, and c = 14.
– b  b 2 – 4( a )( c )
2( a )
x = 1
Substitute values in the quadratic formula.
The equation has two solutions:
–9 + 5 2
x =
= –2
–9  – 56 2
x =
Simplify.
–9  5 2
x =
–9 – 5 2
x =
= –7
–9  2
x =
Simplify.

16. 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
x 2 + 9x + 14 = 0
(–2) 2 + 9(–2) + 14 = 0 ?
(–7) 2 + 9(–7) + 14 = 0 ?
4 + – = 0 ?
49 + – = 0 ?
0 = 0
0 = 0

17. Finding the x-Intercepts of a Graph
Find the x-intercepts of the graph of y = –x 2 – 2 x + 5.
SOLUTION
The x-intercepts occur when y = 0.
y = –x 2 – 2 x + 5
Write original equation.
Substitute 0 for y, and
identify a = 1, b = –2, and c = 5.
0 = –1x 2 – 2 x + 5
The equation has two solutions:
–( b )  ( b ) 2 – 4( a )( c )
2( a )
x =
– – –1 5 –1
Substitute values in the quadratic formula. –2
x =
 –3.45
2  –2
x =
2  24 –2
x =
Simplify.
2 – 24 –2
x =
2  24 –2
x =
 1.45
Solutions

18. Finding the x-Intercepts of a Graph
Check your solutions to the equation y = –x 2 – 2 x + 5 graphically.
Check y  –3.45 and y  1.45.
You can see that the graph shows the x-intercepts between –3 and –4 and between 1 and 2.

19. 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
OBJECT IS THROWN:
h = –16 t 2 + v t + s
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.

20. Modeling Vertical Motion
BALLOON COMPETITION
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?
SOLUTION
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.

21. –16 Modeling Vertical Motion BALLOON COMPETITION SOLUTION
v = –30, s = 200, h = 0.
h = –16 t 2 + v t + s
Choose the vertical motion model for a thrown object.
Substitute values for v and s into the vertical motion model.
h = –16 t v t + s
(–30)
h = –16 t 2 – 30t + 200
Substitute 0 for h. Write in standard form.
–( b )  ( b ) 2 – 4( a )( c )
2( a )
t =
– – –
–16
Substitute values in the quadratic formula.
30  13,700
–32
t =
Simplify.
t  2.72 or –4.60
Solutions

22. Modeling Vertical Motion
BALLOON COMPETITION
SOLUTION
t  2.72 or –4.60
As a solution, –4.60 does not make sense in the context of the problem.
Therefore, the weighted marker will hit the target about 2.72 seconds after it was thrown.

23. Solving Quadratic Equations by the Quadratic Formula
Try the following examples. Do your work on your paper and then check your answers.
Skip #4 and #5

24. Homework
Homework – P. 245 #12,15,18,21,23,24,25,28,37,38