Essential Question: How do we do this stuff?

1) Use the x-intercept method to find all real solutions of the equationx3 – 8x2 + 9x + 18 = 0

Graph the function using the graphing calculator

Find the roots Roots at -1, 3, & 6

2) Determine the nature of the roots2x2 – 12x + 18 = 0

Use the discriminant to determine the number of roots:

Discriminant = 0 means “1 real solution”

2

2

4

( 12) 4(2)(18)

144 144

0

b ac

3) Solve by taking the square root of both sides: (4x-4)2 = 25

2(4 4) 25

4 4 5

4 4 5 or 4 4 5

4 9 or

9 1 or

4

4 4

4

4 1

x

x

x

x

x x

x

x

4) Solve by factoring: x2 + 2x – 3 = 0◦ Looking for two numbers that multiply to get -3

and add to get 2◦ Only ways to multiply to get -3 are

1 • -3 (they add to -2) -1 • 3 (they add to 2) Hey! We got a winner!

◦ Factor using those numbers (x – 1)(x + 3) = 0

◦ Set each part of the factorization to 0 to get the solutions x – 1 = 0 or x + 3 = 0 x = 1 or x = -3

5) Solve by using the quadratic formulax2 – 2x – 5 = 0 2

2

4

2

( 2) ( 2) 4(1)( 5)

2(1)

2 4 20 2 24

2

1 6

2

2 4 6 2 2 6

2 2

b b ac

a

6) Find all solutions: 5x = 2x2 - 12

2

2

5 2 1

0 2 5 1

( 5) (

5 33

5 5

5) 4(2)( 1)

2(2)

5 25 8

4

4

x x

x

x x

x

7) Find all solutions: |4 – 0.2x| + 1 = 19

4 0.2 1 19

4 0.2 18

4 0.2 18 or 4 0.2 18

4 0

70 or 110

.2 18 or 4 0.2

1 1

4 4 4 418

0. 0.2 0.2 0.22 14 0.2 or 0.2 22

x

x

x x

x x

x x

x x

8) Find all solutions: |x2 - 10x + 17| = 82 2

2 2

2 2

10 17 8 or 10 17 8

10 17 8 or 10 17 8

10 9 0 or 10

9, 1

25 0

( 9)( 1) 0 or ( 5)(

8

5)

8 8 8

o 5

0

r x

x x x x

x x x x

x x x x

x x x x

x x

9) Find all solutions: 3 3 4 7 5x

3

33 3

3 4 7 5

3

7 7

4 4

3 3

4 2

3 4 8

3 4

4

3

x

x

x

x

x

10) Find all solutions: 2 6 21 4x x

2

2

2

2

2

6 21 4

6 21 16

6

16 16

5 or

5 0

( 1)

5)( 0

1

x x

x x

x x

x

x x

x

11) The problem on the preview has no solution (square roots can’t ever be negative)Find all solutions: 0 71x x

2 2

22

10 7

10 7

10 49 14

39 14

49 49

1961521 19

1521

1

19

6

6

96

x x

x x

x

x x

x xx

x

x

x

x

12) Find all solutions:

◦ Real solutions? When numerator = 0 x2 + 1x – 42 = 0 (x + 7)(x – 6) = 0 x = -7 or x = 6

◦ Extraneous solutions? When denominator = 0 x – 6 = 0 x = 6

◦ When a solution comes up as real and extraneous, the extraneous solution takes precedence Real solution: x = -7 Extraneous solution: x = 6

2 1 420

6

x x

x

13) Find all solutions:

◦ Real solutions? When numerator = 0 5x2 + 44x + 63 = 0 (5x + 9)(x + 7) = 0 x = -9/5 or x = -7

◦ Extraneous solutions? When denominator = 0 x2 + 12x + 35 = 0 (x + 7)(x + 5) = 0 x = -7 or x = -5

◦ When a solution comes up as real and extraneous, the extraneous solution takes precedence Real solution: x = -9/5 Extraneous solution: x = -7 or x = -5

2

2

5 44 630

12 35

x x

x x

14) Write -4 < x < 9 in interval notation

If an inequality has a line underneath it, we use braces; parenthesis without.

(-4, 9]

15) Solve the inequality and express your answer in interval notation: 2x – 6 < 3x + 8

[-14, ∞)

3 3

6

2 6 3 8

6 8

14

1

6

1 1

4

x xx

x

x

x

x

16) Solve the inequality and express your answer in interval notation: -15<-3x+3<-3

[2, 6]

3 3 3

3

15 3 3 3

18 3 6

6 2

2 6

3 3

x

x

x

x

17) Solve the inequality and express your answer in interval notation:

Critical Points◦ Real solutions: 5 & -9◦ Extraneous solution: 1

Test the intervals◦ (-∞, -9] use x = -10, get -15/11 > 0 FAIL ◦ [-9, 1) use x = 0, get 45 > 0 PASS◦ (1, 5] use x = 2, get -33 > 0 FAIL◦ [5, ∞) use x = 6, get 3 > 0 PASS

Interval solutions are [-9, 1) and [5, ∞)

( 5)( 9)0

( 1)

x x

x

18) The simple interest I on an investment of P dollars at an interest rate r for t years is given by I = Prt. Find the time it would take to earn $1800 in interest on an investment of $17,000 at a rate of 6.9%.

You’re given I ($1800), P ($17,000) and r (6.9% = 0.069).

Just plug them into the equation and solve for t

◦ 1800 / 17000 = (17000)(0.069)(t) / 17000◦ 0.10588 / 0.069 = (0.069)(t) / 0.069◦ 1.53 = t

19) d = -16t2 + 37. Find how long it takes the object to reach the ground (d = 0)

Because time is never negative, t = 1.5 s

2

2

2

2

37 37

0 16 37

0 16 37

37 16

2.3125

1.5

t

t

t

t

t

20) 128t – 16t2. During what period of time is the arrow above 240 feet

2

2

22

2 2128 16 240

0 16 128 240

( 128) ( 128) 4(16)(240)4

2 2(16)

128 16384 15360 128 1024

32 32128 32 128 32 160 128 32 96

or 32 32 32 32 32

128 1

3 or

6 128 16

5

t tt t

t t

b b

t

x

t

ac

a

x

#20, continued◦ 16t2-128t+240 < 0

Test the intervals◦ (-∞, 3] -> test x = 0, get 240 < 0 FAIL◦ [3, 5] -> test x = 4, get -16 < 0 PASS◦ [5, ∞) -> test x = 6, get 48 < 0 FAIL

The arrow is above 240 ft. from 3 to 5 sec.