Download Name __________________________________________ DUE: February 28, 2011 Algebra II – Pd ____

Name __________________________________________
Algebra II – Pd ____
DUE: February 28, 2011
Inequalities Packet
1. Absolute Value Inequalities
To solve you must:
1. Treat the inequality sign as an equal sign and solve as you would an absolute value equation.
2. Once you have your two answers, place them in numerical order on a number line.
3. Above the 2 numbers on the number line put
a. an “Open Circle” if the question is < or >
b. a “Closed Circle” if the question is < or >
3. Chose a “Test Point” in each section and test that number in the original problem to see if
it is TRUE or FALSE
4. SHADE in the “TRUE” Sections
5. Write an algebraic Solution Set
Solution Set : {-4 < x < 0}
Example: Solve
|x + 2| < 2
x+2=2
x=0
x + 2 = -2
x = -4
-4
Test : x = -5
|x + 2| < 2
|(-5) + 2| < 2
|-3| < 2
3<2
False
Solution Set :
Example: Solve
2|x - 5| > 8
2
2
|x - 5| = 4
x-5=4
x=9
x - 5 = -4
x=1
x = -3
|x + 2| < 2
|(-3) + 2| < 2
|-1| < 2
1<2
True
Test : x = 0
2|x - 5| > 8
2|(0) - 5| > 8
2|-5| >8
2(5) > 8
10 > 8
True
Example: Solve
9
x=3
2|x - 5| > 8
2|(3) - 5| > 8
2 |-2| > 8
2 (2) > 8
4>8
False
Test : x = -3
4 – 4|8x – 4| > -76
4 – 4|8(-3) – 4| > -76
4 – 4|-24 – 4| > -76
4 – 4|-28| > -76
4 – 4(28) > -76
4 – 112 > -76
True
3
x=0
4 – 4|8x – 4| > -76
4 – 4|8(0) – 4| > -76
4 – 4|0 – 4| > -76
4 – 4|– 4| > -76
4 – 4(4) > -76
4 – 16 > -76
-12 > -76
False
1
x = 10
2|x - 5| > 8
2|(10) - 5| > 8
2|5| > 8
2 (5) > 8
10 > 8
True
{x < -2 ∪ 𝑥 > 3}
-2
-108 > -76
x=1
|x + 2| < 2
|(2) + 2| < 2
|4| < 2
4<2
False
{x < 1 ∪ 𝑥 ≥ 9}
1
Solution Set :
4 – 4|8x – 4| > -76
–4
–4
– 4|8x – 4| = -80
-4
-4
|8x – 4| = 20
8x - 4 = 20
8x - 4 = -20
8x = 24
8x = -16
x=3
x = -2
0
x=4
4 – 4|8x – 4| > -76
4 – 4|8(4) – 4| > -76
4 – 4|32 – 4| > -76
4 – 4|28| > -76
4 – 4(28) > -76
4 – 112 > -76
-108 > -76
True
Name __________________________________________
Algebra II – Pd ____
Practice:
1. |-6 + x| > 12
2. |x + 1| < 1
3. 5 + |x + 1| < 8
2
DUE: February 28, 2011
Inequalities Packet
Name __________________________________________
Algebra II – Pd ____
4. |-3x| - 6 > -3
5. 3|x + 5| < 6
6. |8 + 9x| + 3 > 56
3
DUE: February 28, 2011
Inequalities Packet
Name __________________________________________
Algebra II – Pd ____
7. 3 + 2|9 + x| > -1
8. 10|x + 6| -7 > 73
4
DUE: February 28, 2011
Inequalities Packet
Name __________________________________________
Algebra II – Pd ____
9. 5|5x + 7| + 3 > 18
10. -4| -3 + 7x| + 9 < -59
5
DUE: February 28, 2011
Inequalities Packet
Name __________________________________________
Algebra II – Pd ____
DUE: February 28, 2011
Inequalities Packet
2. Quadratic Inequalities
To solve you must:
1. Treat the inequality as an equal sign and set “equal” to zero.
2. Solve to find the roots of the equation by factoring (or use another method)
3. Place both roots on a number line, in numerical order
4. Above the 2 numbers on the number line put
a. an “Open Circle” if the question is < or >
b. a “Closed Circle” if the question is < or >
5. Chose a “Test Point” in each section and test that number in the original problem to see if
it is TRUE or FALSE
6. Write an algebraic Solution Set
Solution Set : {-8 < x < 6}
Example: Solve
x2 + 2x – 48 < 0
x2 + 2x – 48 = 0
(x + 8)(x – 6) = 0
x = -8
x=6
-8
Test : x = -9
6
x=0
x2 + 2x – 48 < 0
2
(-9) + 2(-9) – 48 < 0
81 – 18– 48 < 0
15 < 0
False
x=7
x2 + 2x – 48 < 0
(0)2 + 2(0) – 48 < 0
0 – 0 – 48 < 0
-48 < 0
True
x2 + 2x – 48 < 0
(7)2 + 2(7) – 48 < 0
49 + 14 – 48 < 0
15 < 0
False
Solution Set : {-2 < x < 2}
2x2 -26 < -7x2 + 10
2x2 -26 = -7x2 + 10
+7x2 – 10 + 7x2 – 10
9x2 – 36 = 0
9(x2 – 4) = 0
9(x – 2)(x + 2) = 0
9≠0 x = 2 x = -2
Example: Solve
-2
2
Test : x = -3
x=0
2x2 -26 < -7x2 + 10
2x2 -26 < -7x2 + 10
2(-3)2 -26 < -7(-3)2 + 10 2(0)2 -26 < -7x(0)2 + 10
2(9) -26 < -7(9) + 10
2(0) -26 < -7(0) + 10
18 – 26 < -63 +10
0 – 26 < 0 +10
-8 < -53
- 26 < 10
False
True
1
x=3
2x2 -26 < -7x2 + 10
2(3)2 -26 < -7(3)2 + 10
2(9) -26 < -7(9) + 10
18 – 26 < -63 +10
-8 < -53
False
2
Solution Set : {x < − 2 ∪ 𝑥 ≥ 3}
6x2 – x – 2 > 0
6x2 – x – 2 = 0
2
6x – 4x + 3x – 2 = 0
2x
+1
2x(3x – 2) +1(3x -2) = 0
(3x – 2)(2x + 1) = 0
2
1
x=
x=−
−
Example: Solve
3
2
Test : x = -1
6x2 – x – 2 > 0
6(-1)2 – (-1) – 2 > 0
6(1) + 1 – 2 > 0
6+1–2>0
5>0
True
6
1
2
2
( )
3
x=0
6x2 – x – 2 > 0
6(0)2 – (0) – 2 > 0
6(0) – 0 – 2 > 0
0–0–2>0
–2>0
False
x=1
6x2 – x – 2 > 0
6(1)2 – (1) – 2 > 0
6(1) –1 – 2 > 0
6–1–2>0
3>0
True
Name __________________________________________
Algebra II – Pd ____
11. x2 + 6x – 16 < 0
12. x2 + 2x – 8 < 0
13. x2 – 4 > 0
7
DUE: February 28, 2011
Inequalities Packet
Name __________________________________________
Algebra II – Pd ____
14. 2x2 + 4x > x2 – x – 6
15. 5x2 + 2x < 0
16. x2 + 2x > 15
8
DUE: February 28, 2011
Inequalities Packet
Name __________________________________________
Algebra II – Pd ____
17. 2x2 + 14x < x2 – 45
18. 9x2 – 3x < 2
9
DUE: February 28, 2011
Inequalities Packet
Name __________________________________________
Algebra II – Pd ____
19. x2 + 71 ≥ -18x – 9
20. 4x2 < 256
10
DUE: February 28, 2011
Inequalities Packet