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