# Day 4 Problems 1 5 07 • Solve each equation 1 1 4 x  1  3 2 4  7  y 2  0 3 1  x  2  0 4 z6 30 5 3 by anf21181

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Cube Root Worksheet document sample

• pg 1
```									             Day 4 Problems
1/5/07
• Solve each equation.                1
1. 4 x  1  3        2. 4  (7  y ) 2  0

3. 1  x  2  0       4.    z6 30

5.   3
a4 3
Cube Root Equation
1

• Solve       3(5n  1)  2  0 .
3
1
3(5n  1)  2  0
3
1
3(5n  1)  2    3
35
1
2                5n 
(5n  1) 
3
27
3
1 3                   7
          2
3

(5n  1)    
3                n
           3                 27
8
5n  1 
27
• MAKE SURE YOU CHECK YOUR SOLUTION!!!!
Cube Root Equation
1

• CHECK!           3(5n  1)  2  0
3

1 ?
7
3(5     1)  2  3
27
1
 8   3?
3   2  0
 27 
2     ?
3   2  0
 3
00                  7
• The solution checks, so 27   is the solution.
Inequalities Cont.
inequality that has a variable in
–Ex. x  2  4
• Solve 2  4 x  4  6 .
Since the radicand of a square root must be
greater than or equal to zero, first solve
4x – 4 ≥ 0.
4x – 4 ≥ 0
4x ≥ 4
x≥1
• Now solve 2  4 x  4  6.

2  4x  4  6

4x  4  4
4 x  4  16
4 x  20
x5
It appears that 1  x  5. You can test
some x values to confirm the solution.
• Let    f ( x )  2  4 x  4.
x=0                  x=2                  x=7

f (0)  2  4(0)  4 f (2)  2  4(2)  4 f (7)  2  4(7)  4

 2  4            4                   6.90

Since 4 is not a Since 4  6 ,     Since 6.90  6 ,
real number, the  the inequality is the inequality is
inequality is not satisfied.        not satisfied.
satisfied.

So, the solution checks. You can summarize the
solution with a number line.
1                5
Concept Summary
• To solve radical inequalities, complete the
following steps.
– Step 1 – If the index of the root is even,
identify the values of the variable for which