Review 4.6-4.7
Solve each equation or inequality
1.
€
b −10
b= −3 2.
€
3y +1
4+
2 + 4y
3= −
5
6
b = 2, -5
€
y =−21
25
Multiply every term by 12
€
3(3y +1) + 4(2 + 4y) = 2(−5)
9y + 3+ 8 +16y = −10
25y = −21
€
b −10
b= −3
⎡ ⎣ ⎢
⎤ ⎦ ⎥(b)
b2 −10 = −3b
b2 + 3b −10 = 0
Solve each equation or inequality
3.
€
2
w+
6
w −1≤ −5 4.
€
a
a − 2+
6
a+ 2= 2
€
−1 ≤ w < 0
25 ≤ w <1
€
4 ± 2 3
€
€
2
w+
6
w −1≤ −5
⎡ ⎣ ⎢
⎤ ⎦ ⎥(w)(w −1)
€
2(w −1) + 6(w) ≤ −5(w)(w −1)
2w − 2 + 6w + 5w2 − 5w ≤ 0
5w2 + 3w − 2 ≤ 0w = 2/5, -1
€
a
a − 2+
6
a+ 2= 2
⎡ ⎣ ⎢
⎤ ⎦ ⎥(a − 2)(a+ 2)
€
a(a+ 2) + 6(a − 2) = 2(a2 − 4)
a2 + 2a+ 6a −12 − 2a2 + 8 = 0
−a2 + 8a − 4 = 0
USE QUADRATIC FORMULA
€
−8 ± 64 − 4(−1)(−4)
−2What values go on your number lines??
2/5, -1, 0, and 1
€
−8 ± 48
−2=
−8 ± 4 3
−2
Example: Solve.
xx
12
2
13
LCM: 2x
Multiply each fraction through by
the LCM
x
xx
x
x 12*2
2
23*2
246 x
18 x
18xCheck your solution!
18
12
2
1
18
3
1293
Solve.1
54
1
5
xx
x LCM: ?LCM: (x+1)
)1(
)1(5)1(4
)1(
)1(5
x
xx
x
xx
5445 xx
145 xx
1x
Check your solution!
11
54
11
)1(5
0
54
0
5
?
No Solution!
Solve. 14
6
2
232
xx
x
Factor 1st!
1)2)(2(
6
2
23
xxx
x
LCM: (x+2)(x-2)
)2)(2()2)(2(
)2)(2(6
)2(
)2)(2)(23(
xx
xx
xx
x
xxx
42264263 22 xxxxxx
2443 22 xxx
0642 2 xx0322 xx
0)1)(3( xx
01or 03 xx
1or 3 xx
Check your solutions!
Example: Solve.
4
1
4
32
xxx
xx 42 12342 xxx
0122 xx0)3)(4( xx
03or 04 xx3or 4 xx
Check your solutions!
)4(3 x
Last Example: Solve. 1
2
22
62
x
x
xx
)1(6 x)2)(1(2 xxx
6)2(2 xx3)2( xx
0322 xx0)1)(3( xx
01or 03 xx1or 3 xx
1
2
)1(2
6
x
x
xx
Check your solutions!
Solve Check your solution.
The LCD for the three denominators is
Original equation
Multiply each sideby 24(3 – x).
1 1
11 1
6
Simplify.
Simplify.
Add.
Check Original equation
Simplify.
Simplify.
The solution is correct.
Answer: The solution is –45.
Answer:
Solve
Solve Check your solution.
The LCD is
Original equation
Multiply by the LCD, (p2 – 1).
p – 1
1
1
1
DistributiveProperty
Simplify.
Simplify.
Add(2p2 – 2p + 1)to each side.
Factor.
or Zero ProductProperty
Solve eachequation.
Divide eachside by 3.
Check Original equation
Simplify.
Simplify.
Since p = –1 results in a zero in the denominator, eliminate –1.
Answer: The solution is p = 2.
Simplify.
Original equation
Answer:
Solve
Decompose this into partial fractions
5.
€
4 p2 +13p −12
p3 − p2 − 2p
€
6
p+
5
p − 2−
7
p+1
Solve
Step 1 Values that make the denominator equal to 0 are excluded from the denominator. For this inequality the excluded value is 0.
Step 2 Solve the related equation.
Related equation
Multiply each side by 9s.
Simplify.
Add.
Divide each side by 6.
Step 3 Draw vertical lines at the excluded value and at the solution to separate the number line into regions.
Now test a sample value in each region to determine if the values in the region satisfy the inequality.
Test
is a solution.
is not a solution.
Test
is a solution.
Test
Answer: The solution
Solve
Answer:
Solve each equation or inequality
6.
€
x − 3 + 4 = 6 7.
€
2x − 3 = 5x + 4
x = 7 No real solution
Solve each equation or inequality
8.
€
k − 7 + k − 3 = 2 9.
€
2m −13 + 6 = 3
k = 7 m = -13
Solve each equation or inequality
10.
€
3t + 7 > 7
t > 14
11.) Find the upper and lower bound of the zeros of:
€
f (x) = x 3 − 3x 2 − 2x −1
Upper = 4Lower = -1
Approximate the real zeros of each functions
12.
€
f (x) = −x 3 − 2x + 4 13.
€
f (x) = x 4 − 3x 3 + 2x −1
Between 1 and 2 Between -1 and 0Between 2 and 3
Approximate the real zeros of each functions
14.
€
f (x) = x 3 + 4x 2 − 3x − 5 15.
€
f (x) = −3x 4 − 5x 3 + x − 2
Between -5 and -4, -1 and 0, 1 and 2 No real zeros
16.) Find the number of positive, negative, and imaginary:
€
f (x) = x 3 − x 2 + 6x +1
Pos. 2 or 0Neg. 1
17.) Find the possible rational roots:
€
f (x) = 2x 3 − 4x 2 + 5x − 3
€
±1,±3,±1.5,±0.5
Find the remainder of each division.Then state whether the binomial is a factor?18.
€
(x 3 −14x) ÷ (x − 5) 19.
€
x 3 − 6x + 9
x − 3
Remainder = 55Not a factor
Remainder = 18Not a factor
20.) Use synthetic division to divide:
€
x 3 − 5x 2 −17x − 6
x + 2
€
x 2 − 7x − 3
Graph each equation or inequality:
21.
€
y < 4 − x − x 2 22.
€
y = x 2 + 4x + 4
Solve each equation or inequality23.
€
2x 2 − 5x + 2 = 0 24.
€
3x 2 − x +10 = 0
€
x = 2
x = 12
€
x =1± i 119
6
Solve each equation by completing the square:
25.
€
x 2 − 5x − 84 = 0 26.
€
x 2 −7
12x +
1
12= 0
€
x =12
x = −7
€
x =1
3
x =1
4
27.) Find the discriminant of the function given and describe the nature of the roots.
Discriminant = 732 distinct real roots
€
2x 2 − 8 + 3x = 0
Write the polynomial equation of least degree have the roots:
28.) 1, -1, 0.5
€
2x 3 − x 2 − 2x +1 = 0
Find the roots of each equation:
29.
€
x 2 + 36 = 0 30.
€
4x 3 −10x 2 − 24x = 0
€
x = ±6i
€
x = 0,4,−1.5