### Coming up in Math 110:

• • • • Today: Section 8.2 (Quadratic formula) Tomorrow: Review for Quiz 4.

NOTE:

Practice Quiz 4 is open now and ready for you to take as many times as you want.

Wednesday: Quiz 4 on sections 5.5 - 5.8 & 8.2 (Factoring polynomials and solving polynomial equations.) Thursday: Review for Test 2

## Monday: Test 2

(100-point test on all of Chapter 5 + section 8.2)

NOTE:

Because this test occurs on a Monday, there will be an

optional review session on Sunday, March 30 th

. That session will be held from

7:00-9:00 p.m

. in this classroom (214) and it is open to all Math 110 students from all sections.

## Any questions on the Section 5.8 homework?

### Pass your worksheets for this assignment to the middle aisle for pickup now.

Remember the problem like this one from the homework that was due today?

Wouldn’t it be nice if there was an easier way to do it than by factoring?

Leave factoring up on board: (4x - 9)(3x + 8)

## Section 8.2

The

is another technique we can use for solving

.

Remember,

are

polynomial

equations of

degree 2

, such as x 2 + 3x -7 = 0 or 5x 2 – 14 = 0 .

The quadratic formula is derived from a process called “completing the square” for a general quadratic equation. –

See Section 8.1 if you’re interested in seeing how this formula is derived.

This will also be covered in Math 120 in more detail, along with the technique called “completing the square”.

x 2

x

x

 

b

b

2 

ac

### 2

a

Note:

This formula

IS

on the pink formula sheet, but you’ll probably have it memorized by the time you’ve done the first few homework problems.

# The Big Question:

### ?

Example 1

Solve x 2 + 4x + 3 = 0 by • • Factoring The quadratic formula.

Which way works best?

2

### + 4x + 3 = 0 by Factoring: This one is pretty easy to factor. The factoring is (x + 3)(x + 1) = 0 , so the solutions are given by x + 3 = 0 , or x = -3 , and x + 1 = 0 , which gives x = -1 .

Now, solve x 2 + 4x + 3 = 0 by the quadratic formula: a = 1, b = 4, c = 3, so the formula gives:

x

or

   4 4    4 2  2 2 4  4 2 2  1  4  1  3   4  2  2   2 4   4  2  6   3 2 16  2   2 2 12   1

### Which way works best for this problem?

In this case, the

factoring method

is much quicker, although BOTH methods give the same answer .

Example 2

Solve x 2 + 5x + 12 = 0 by • • Factoring The quadratic formula.

Which way works best?

Solve x 2 + 5x + 12 = 0 by Factoring: This one

looks

pretty easy to factor, but when you start trying to find two factors of 12 that add up to 5, nothing works. (1+12=13, 2+6=8, 3+4=7). What does this mean? It means that the polynomial is

PRIME

, and there are no rational solutions. (Remember, a rational number is either an integer or a fraction.)

Solve x 2 + 5x + 12 = 0

(continued):

• Let’s see what the quadratic formula gives in this case: a = 1, b = 5, c = 12 so the formula gives:

x

  5  5 2  2  1 4  1  12   5  2 25  48   5  2  23 Notice that the number under the radical sign is negative, which means there are no

real

answers. If the number under the square root sign comes out to be positive but it’s not a perfect square, this means the answer

is

a real number, but is

irrational

because it can’t be simplified to remove the radical. In either of these cases, we’d say the polynomial is

prime

, and therefore has

no rational roots

.

2

### ?

Either way works fine, but if you think a polynomial is prime, a good way to check is by calculating the

discriminant (b 2 – 4ac).

If the discriminant is either

negative

or

not a perfect square

, then you know for sure that your polynomial is prime and there are

no rational

solutions.

Example 3

Solve x 2 + 2x - 4 = 0 by • • Factoring The quadratic formula.

Which way works best?

Solve x 2 + 2x - 4 = 0 by Factoring: Again, this one

looks

like it would be easy to factor, but when you start trying to find two factors of 4 that add up to

2

, nothing works. (-1 + 4 =

3

, -4 + 1=

-3

, -2+2 =

0

) What does this mean? It means that the polynomial is

PRIME

, and there are no rational solutions. (Remember, a rational number is either an integer or a fraction.)

Solve x 2 + 2x - 4 = 0

(continued):

• Let’s see what the quadratic formula gives in this case: a = 1, b = 2, c = -4 so the formula gives:

x

  2  2 2  2  1 4  1 (  4 )   2  2 4  16   2  2 20 Notice that the number (20) under the radical sign is positive, which means there are two

real

answers. However, since 20 is not a perfect square, its square root is

irrational

. In this case, we’d again say that the polynomial is

prime

, because it has

no rational roots

.

Solve x 2 + 2x - 4 = 0 using the quadratic formula

(continued): x

  2  2 2  2  1 4  1 (  4 )   2  2 4  16   2  2 20

IMPORTANT NOTE

: Because the discriminant (20) is POSITIVE, we still do have two REAL NUMBER solutions, but they won’t be rational numbers. How do we find these two real answers if we can’t factor the polynomial? Answer: continue calculating the answer using the quadratic formula, So, if you were asked to find the approximate solution(s) of this equation rounded to the nearest hundredth, you’d use your your two answers.

Answers: x = 1.24 and x = -3.24.

(

How would you check this?

Will you get EXACTLY zero?)

Now re-do this problem from the 5.8 homework using the quadratic formula: Do this on your calculator now, showing work in your notebook.

• What is the discriminant?

• • • • Is it positive? Is it a perfect square? How many real answers will there be? Will they be rational or irrational?

### Recap:

For the equation 12x 2 + 5x -72 = 0, should we solve it using the factoring approach or the quadratic formula? Either way works, but the quadratic formula approach is probably going to be faster than factoring for most people.

Moral of the story

:

,

it’s probably going to be quicker to solve it using the quadratic formula than it would be to factor the polynomial.

Question:

Clear them first by multiplying all terms by the LCD: 1 Solve

x

2 8

x

2 +

x

+ 8

x

– 5 = 0 by the quadratic formula.

2 – 20 = 0 (multiply both sides by 8)

a

= 1,

b

= 8,

c

= -20

x

  8  ( 8 ) 2  4 ( 1 )(  20 )   8  2 ( 1 ) 64  80 2   8  2 144   8  12  2  20 or 2 4 2 ,  10 or 2

b

2 – 4

ac

) is called the

discriminant

.

• The discriminant will take on a value that is positive, 0, or negative.

• The value of the discriminant indicates two distinct real solutions (if it’s positive), one real solution (if it’s zero), or two complex, but not real solutions (if it’s negative – a topic to be discussed in Math 120).

The Discriminant and the Kinds of Solutions to ax 2 + bx +c = 0 Discriminant

b

2 – 4

ac

Kinds of solutions to

ax 2

+

bx

+

c =

0 Graph of

y

=

ax 2

+

bx

+

c b

2 – 4

ac > 0 b

2 – 4

ac = 0

Two unequal real solutions (If

b 2 – 4ac

is a perfect square, the two solutions will be rational numbers. If not, they’re both irrational.) One real solution (a repeated solution) (If

b 2 – 4ac

is a perfect square, the solution will be a rational number. If not, it’s irrational.) Two

x

-intercepts One

x

-intercept

b

2 – 4

ac < 0

No real solution; two complex imaginary solutions No

x

-intercepts

Example

Use the discriminant to determine the

number

and

type

of solutions for the following equation.

5 – 4

x

+ 12

x

2 = 0

b

2 – 4

ac a

= 12,

b

= (-4) 2 = -4, and

c

– 4(12)(5) = 5 = 16 – 240 = -224 Since the discriminant is negative , there are no real solutions.

Question:

What would this graph look like?

Example

Use the discriminant to determine the

number

and

type

of solutions for the following equation.

25x 2 - 4 = 0

b

2 – 4

a ac

= 25 , = (0) 2

b

= 0

(why?)

, and

c

= -4 – 4(25)(-4) = 0 – -400 = 400 Since the discriminant is positive , there are two real solutions. (You could go on to show that the solutions are 2/5 and -2/5, either by factoring or using the quadratic formula.)

Example

Use the discriminant to determine the number and type of solutions for the following equation.

5 – 4

x

+ 12

x

2 = 0

a

= 12,

b

= -4, and

c

= 5

b

2 – 4

ac

= (-4) 2 – 4(12)(5) = 16 – 24 = -224 Since the discriminant is negative , there are no real solutions.

NOTE:

There will be two

COMPLEX

solutions that are not real numbers, i.e. they will contain the imaginary number

i

which is the square root of -1.

Example

Use the discriminant to determine the

number

and

type

of solutions for the following equation.

x 2 – 8x + 16 = 0

a

= 1,

b

= -8, and

c

= 16

b

2 – 4

ac

= (-8) 2 – 4(1)(16) = 64 – 64 = 0 Since the discriminant is zero , there is

one

real solution. (You could go on to show that the solution is 4, either by factoring or using the quadratic formula.)

Question:

What would this graph look like?

How do you figure out the answers if the discriminant is positive but not a perfect square?

2 possible approaches: 1. Exact

(More on this in Chapter 7...)

2. Approximate

### REMINDER!!!

IMPORTANT NOTE:

Use the

technique to solve

all

problems in this homework assignment.

There are a couple of word problems at the end of the assignment in which the online learning aids will show factoring as the solution method. You should use the quadratic formula instead (and you will find it to be easier and quicker than factoring.)

OPEN

### your LAPTOPS and begin working on the homework assignment

(if there’s any time left...) But remember, you can always work in the JHSW 203 open lab after class (or before tour next class session) if you want some help on this homework.