Download Name __________________________________________ DUE: February 23, 2015 Algebra II – Pd ____

Name __________________________________________
Algebra II – Pd ____
DUE: February 23, 2015
Inequalities Packet
Inequalities Packet
This assignment is to be completed by you over February break. You are to record your answers on this
sheet of paper. ALL OF YOUR WORK MUST BE WRITTEN NEATLY IN THE PACKET. If your
work is not shown for every problem, you will receive no credit for this assignment. This assignment will
be counted as a quiz and you will also take an in-class quiz on this material after the break. This assignment
is due on Monday February 23, 2015. IN CLASS. Again, this assignment is due in class, NOT in my
mailbox, NOT after school, NOT during lunch or you will receive a zero! If you need to ask any questions,
please feel free to attend extra help during school days, the math center during your frees. I will not be
available for help during the break Please make sure you look at the packet before the break. If you hand in
your Packet before break, you will receive 5 bonus points.
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1
Name __________________________________________
Algebra II – Pd ____
DUE: February 23, 2015
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
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
x = -3
|x + 2| < 2
|(-3) + 2| < 2
|-1| < 2
1<2
True
{x < 1
}
1
9
Test : x = 0
2|x - 5| > 8
2|(0) - 5| > 8
2|-5| >8
2(5) > 8
10 > 8
True
Solution Set :
Example: Solve
0
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
False
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
True
2
x = 10
2|x - 5| > 8
2|(10) - 5| > 8
2|5| > 8
2 (5) > 8
10 > 8
True
{-2 < x < 3 }
-2
-108 > -76
x=1
|x + 2| < 2
|(2) + 2| < 2
|4| < 2
4<2
False
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
False
Name __________________________________________
Algebra II – Pd ____
Practice:
1. |-6 + x| > 12
2. |x + 1| < 1
3. 5 + |x + 1| < 8
3
DUE: February 23, 2015
Inequalities Packet
Name __________________________________________
Algebra II – Pd ____
4. |-3x| - 6 > -3
5. 3|x + 5| < 6
6. |8 + 9x| + 3 > 56
4
DUE: February 23, 2015
Inequalities Packet
Name __________________________________________
Algebra II – Pd ____
7. -3 + 2|9 + x| > -1
8. 10|x + 6| -7 > 73
5
DUE: February 23, 2015
Inequalities Packet
Name __________________________________________
Algebra II – Pd ____
DUE: February 23, 2015
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
x2 + 2x – 48 < 0
2
(-9) + 2(-9) – 48 < 0
81 – 18– 48 < 0
15 < 0
False
6
x=0
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
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
Solution Set : {x <
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
x=
x=
2
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
}
Example: Solve
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
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 ____
9. x2 + 6x – 16 < 0
10. x2 + 2x – 8 < 0
11. x2 – 4 > 0
7
DUE: February 23, 2015
Inequalities Packet
Name __________________________________________
Algebra II – Pd ____
12. 2x2 + 4x > x2 – x – 6
13. 5x2 + 2x < 0
8
DUE: February 23, 2015
Inequalities Packet
Name __________________________________________
Algebra II – Pd ____
14. x2 + 2x > 15
15. x2 + 71 ≥ -18x – 9
9
DUE: February 23, 2015
Inequalities Packet