1. Absolute Value Equations

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1. Absolute Value Equations
To solve an absolute value equation, you must:
1. Isolate the Absolute Value
2. Split into two equations – Positive and Negative
(one side equals the positive value and the other side equals the negative value)
**when you have a binomial, use parenthesis and remember to distribute
3. Solve each equation for x
4. Check both answers in the original problem.
Examples: Solve
2 + 3|x + 5| = 14
-2
-2
3|x + 5| = 12
3
3
|x + 5| = 4
x+5=4
x = -1
x + 5 = -4 (negate)
x = -9
Now check both answers
2 + 3|x + 5| = 14
2 + 3|x + 5| = 14
2 + 3|(-1) + 5| = 14
2 + 3|(-9) + 5| = 14
2 + 3|4| = 14
2 + 3|-4| =14
2 + 3(4) = 14
2 + 3(4) = 14
2 + 12 = 14
2 + 12 = 14
14 = 14
14 = 14
√
x = {-9, -1}
√
Practice:
1. | x – 1| = 4
Check 1
|x + 6| - 18 = 2x
+ 18 + 18
|x + 6| = 2x + 18
x + 6 = (2x + 18)
x + 6 = 2x + 18
-x
-x
6 = x + 18
-18
-18
-12 = x
x + 6 = -(2x + 18) distribute neg.
x + 6 = -2x -18
-x
-x
6 = -3x – 18
+18
+18
24 = -3x
-3 -3
-8 = x
Now check both answers
|x + 6| - 18 = 2x
|x + 6| - 18 = 2x
|(-12) + 6| - 18 = 2(-12)
|(-8) + 6| - 18 = 2(-8)
|-6| - 18 = -24
|-2| - 18 = -16
(6) – 18 = -24
(2) – 18 = -16
-12 ≠ -24
-16 = -16
x
x = {-8}
√
-12 is extraneous, so x ≠ -12
2. |-6 + a | = 9
Check 2
Check1
Check 2
3. 2|2x – 4| = 86
Check 1
5.
|π‘₯+4|
10
Check 2
7. |x + 6| = 2x
Check 1
Check 2
6. -4|b – 2|-9 = -37
=1
Check 1
4. |-5x| + 4 = -11
Check 2
Check 1
Check 1
Check 2
Check 2
8. |2x + 12| = 7x – 3
9. |2x – 6| - x = 3
Check 1
Check 2
10. |4x + 5| + 3x = 10
Check 1
Check 1
Check 2
Check 2
2. Radical Equations
To solve a radical equation:
1.
2.
3.
4.
Isolate the radical on one side of the equation
Square or both sides of the equation
Solve the remaining equation for x
Check answer in the original problem. (Extraneous Roots)
Model Problem 1: Solve: 3 = √π‘₯ − 1
2
(3)2 = (√π‘₯ − 1)
9=x–1
10 = x
Check: x = 10
3 = √π‘₯ − 1
3 = √10 − 1
3 = √9
3 = 3 √ Check
Model Problem 2: Solve: √3π‘₯ + 1 − 1 = π‘₯ − 4
√3π‘₯ + 1 = π‘₯ − 3
2
(√3π‘₯ + 1) = (π‘₯ − 3)2
3x + 1 = x2 – 6x + 8
0 = x2 – 9x + 8
0 = (x – 8) (x – 1)
(x – 8) = 0 (x – 1) = 0
x=8
x=1
REMEMBER: Squaring a binomial: FOIL
Check: x = 8
√3π‘₯ + 1 − 1 = π‘₯ − 4
√3(8) + 1 − 1 = 8 − 4
√24 + 1 − 1 = 4
√25 − 1 = 4
5–1=4
4 = 4 √ check
Practice: Solve each radical equation and check.
11. √−8 − 2π‘₯ = 4
Check
x=1
√3π‘₯ + 1 − 1 = π‘₯ − 4
√3(1) + 1 − 1 = 1 − 4
√3 + 1 − 1 = −3
√4 − 1 = −3
2 – 1 = -3
-1 = -3 x οƒŸReject!
12. √π‘₯ − 4 = −12
Check
13. √2π‘₯ − 6 = √3π‘₯ − 14
Check
(hint: If a radical is on both
sides of the equation, square
both sides, then solve)
14. √π‘₯ + 1 + 2 = 4
Check
16. π‘₯ = 4 + √2π‘₯ − 8
Check1
15. √8π‘₯ = π‘₯
Check1
17. 𝑦 = √6𝑦 + 16
Check 2
Check1
Check 2
Check 2
18. √2π‘₯ − 7 = π‘₯ − 3
Check1
19. −3 = √37 − 3π‘₯ − π‘₯
Check 2
20. √3π‘₯ − 8 + 1 = 3
Check1
Check 2
Check
3. Solving Equations with Fractional Exponents
To solve you must:
1. Isolate the expression containing the exponent (the base)
2. Raise both sides of the equation to the reciprocal of the exponent (Give both sides a
new exponent: the reciprocal)
3. Solve for x
4. Check in the original
3
Example 1: Solve
54 = 2π‘₯ 2
REMEMBER:
3
2
54 = 2π‘₯
2
2
Fractional Exponents represent
π‘ƒπ‘‚π‘ŠπΈπ‘…
radicals. 𝑅𝑂𝑂𝑇
3
27 = π‘₯ 2
2
3 2
2
(27)3 = ( π‘₯ 2 )3
9=x
2
3
273 = ( √27 )
= 32 = 9
(You can also use your calculator for this step)
−1
EXAMPLE 2: Solve:
5 = 3 + 4π‘Ž 6
a is the base! We must isolate it.
−1
6
−1
6
2 = 4π‘Ž
1
2
= π‘Ž
Raise each side to the reciprocal power
The reciprocal of
1 −6
(2)
=π‘Ž
−1
6
is -6
Negative exponents: flip the base,
exponent becomes positive
(You can also use your calculator for this step)
26 = a
64 = a
5
EXAMPLE 3: Solve:
-3 + (8 − 2π‘₯)4 = 29 The base is (8 – 2x)! Isolate!
5
(8 − 2π‘₯)4 = 32
4
5 5
4
Reciprocal Power!
4
((8 − 2π‘₯) ) = (32)5
5
8 − 2π‘₯ = ( √32)4
8 − 2π‘₯ = 16
−2π‘₯ = 8
π‘₯ = −4
Now, solve for x!
Practice:
3
21. π‘₯ 4 + 6 = 33
−3
22. π‘₯ 2 =
1
729
1
3
23. 11 = 2π‘₯ 2 − 3
24. (𝑛 − 27)2 = 64
3
5
25. 26 = -1 + (27π‘₯)4
26. 3125= (−1 – 18x)3
−3
3
27. 4𝑏 4 + 20 = 52
28. -54 = 10- (x – 10)2
5
29. -5126 = -6 - 5(3π‘₯ + 22)3
3
30. 3646 = 1 + 5(4π‘₯ + 17)2
Name_____________________________________________________
Class_____
Christmas Break Assignment
Algebra II
Equations Packet
This assignment is to be completed by you over Christmas 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 Tuesday January 3, 2011 IN CLASS. (If your class drops, it will be due Wednesday
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 or e-mail me over break. If you hand in your Packet before break, you will receive 5
bonus points.
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