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Solving Radical Equations and Inequalities
Example 1
Solve a Radical Equation
Solve each equation.
a. –10 + c – 1 = 2
–10 +
c – 1 =2
c – 1 = 12
( c – 1 )2 = (12)2
c – 1 = 144
c = 145
Check
–10 +
Original equation
Add 10 to each side to isolate the radical.
Square each side to eliminate the radical.
Find the squares.
Add 1 to each side.
c – 1 =2
?
–10 + 145 – 1
2
2=2 
Original equation
Replace c with 145.
Simplify.
The solution checks. The solution is 2.
b. 5 +
x – 4 = 2.
5+
x – 4 =2
x – 4 = –3
( x – 4 )2 = (–3)2
x–4=9
x = 13
Check
5+
Original equation
Subtract 5 from each side to isolate the radical.
Square each side.
Find the squares.
Add 4 to each side.
x – 4 =2
?
5 + 13 – 4
5+3
8
?
2
2
2
The solution does not check, so the equation
has no real solution.
The graphing calculator screen shows the
graphs of y = 5 + x 4 and y = 2. The graphs
do not intersect, which confirms that there is no
solution.
Example 2
Cube Root Equation
Solve 6 2 x – 1 – 25 = 5.
3
In order to remove the
1
power, or cube root, you must first isolate it and then raise each side of the
3
equation to the third power.
6 3 2 x – 1 – 25 = 5
6 3 2 x – 1 = 30
3
2x – 1 = 5
3
( 2 x – 1 )3 = (5)3
2x – 1 = 125
2x = 126
x = 63
Check
Original equation
Add 25 to each side.
Divide each side by 6.
Cube each side.
Evaluate the cubes.
Add 1 to each side.
Divide each side by 2.
6 3 2 x – 1 – 25 = 5
?
6 3 2(63) – 1 – 25 5
6 3 125 – 25
?
?
5
6(5) – 25 5
5=5 
The solution is 63.
Original equation
Replace x with 63.
Simplify.
The cube root of 125 is 5.
Subtract.
Example 3 Solve a Radical Inequality
Solve
5x + 4 – 6
2.
Since the radicand of a square root must be
greater than or equal to zero, first solve
5x + 4 0 to identify the values of x for which
the left side of the given inequality is defined.
Now solve 5x
4 –6
5x
4 –6 2
5x
4 8
5x + 4 64
5x 60
x 12
It appears that –
4
x
5x + 4
5x
x
0
–4
–
4
5
2.
Original inequality
Isolate the radical.
Eliminate the radical.
Subtract 4 from each side.
Divide each side by 5.
12. You can test some x values to confirm the solution. Let f(x) =
5x
4 – 6.
5
4
4
5
5
Use three test value: one less than – , one between –
values in a table.
and 12, and one greater than 12. Organize the test
x = –1
4 –6
f(–1) = 5(–1)
= –1 – 6
x=0
4 –6
f(0) = 5(0)
= –4
x = 14
4 –6
f(14) = 5(14)
≈ 2.60
Since –1 is not a real number,
the inequality is not satisfied.
Since –4 2, the inequality is
satisfied.
Since 2.60 2, the inequality is
not satisfied.
The solution checks. Only values in the interval –
the solution with a number line.
4
5
x
12 satisfy the inequality. You can summarize