Download Cube Root Equations Exercises: Solve

Name
Class
Date
10-4
Cube Root Equations
Equations containing radicals can be solved by isolating the radical on one side of the
equation and then raising both sides to the same power that would undo the radical.
What is the solution of the radical equation? 2 3 5 x + 1 − 2 =
10
23 5x + 1 − 2 =
10
23 5x + 1 =
12
3
5x + 1 =
6
( 3 5 x + 1)3 =
63
5x + 1 =
216
5 x = 215
x = 43
Add 2 to each side.
Divide each side by 2.
Cube each side to undo the radical.
Simplify.
Subtract 1 from each side.
Divide each side by 5.
Check the solution in the original equation.
Exercises: Solve
1
2. 33 x = 12
1. x 3 = 8
1
4. (3 x + 4) 3 − 1 =
26
7.
1
( x + 10 )3 − 5 =0
1
5. (9 − x) 3 + 12 =
15
8. 3 2 x − 7 − 2 =
3
3.
3
4x + 3 =
67
1
6. 3 7 x + 15 =
9. 3 5 x + 2 − 3 =
0
(continued)
10-4
Cube Root Equations
If an equation contains an expression raised to a rational power, isolate the
expression and raise both sides of the equation to the reciprocal power.
What is the solution? Check your results. 3 ( x + 1)4 − 13 =
3
3
( x + 1)4 − 13 =
3
3
( x + 1)4 =
16
Add 13 to each side to get the radical alone on
one side of the equal sign.
4
Rewrite the radical expression using rational
exponents. Notice that because the numerator of the
exponent is even, we do not know whether the
value of x + 1 is positive or negative.
( x + 1) 3 =
16
3
3
4 4

4
3
x
1
16
+
=
(
)




Raise each side of the equation to the 3 power.
4
Since the root is even, the expression x + 1 may have either a positive
or negative value, so we must consider each case.
|x + 1| = 8
x + 1 =±8
Solve the absolute value equation.
Subtract 1 from each side of the resulting
equations.
x = 7 or x = – 9
Since we introduced and even root we must check for extraneous solutions.
3
3
(7 + 1)4 − 13
3
3
64 − 13
3
16 − 13
3
( x + 1)4 − 13
3
3
(−9 + 1)4 − 13
3
64 − 13
16 − 13
13 = 13
3
3
3
13 = 13
The solutions are 7 and –9.
Exercises: Solve each equation. Check your solution.
4
10. ( x2 + 3 x + 3) 3 =
1
13.
3 2
x
− 5x + 5 − 3 x =
0
3
11. (5 x − 6) 2 =
x3
14.
3
(3 x + 8)2 =
1
12.
3
3 x2 − 3 x + 1 =
x
2
2
15. (3 x + 2) 3 − (4 x + 9) 3 =
0