essential question: how do we do this stuff?. 1) use the x-intercept method to find all real...
TRANSCRIPT
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.