You must have JavaScript enabled to use this site.
mathleaks.com
mathleaks.com
Start chapters
home
Start
History
history
History
expand_more
{{ item.displayTitle }}
navigate_next
No history yet!
Progress & Statistics
equalizer
Progress
expand_more
Student
navigate_next
Teacher
navigate_next
Expand menu
menu_open
Minimize
filter_list
{{ filterOption.label }}
{{ item.displayTitle }}
{{ item.subject.displayTitle }}
arrow_forward
No results
{{ searchError }}
Choose book
search
cancel
menu_open
home
{{ courseTrack.displayTitle }}
{{ statistics.percent }}%
Sign in to view progress
{{ printedBook.courseTrack.name }}
{{ printedBook.name }}
Get free trial
search
Use offline
Tools
apps
Login
account_circle
menu_open
Solving Radical Equations
Choose Course
Algebra 2
Radical Functions
Solving Radical Equations
expand_more
close
Solving Radical Equations 1.3 - Solution
arrow_back
Return to Solving Radical Equations
Solving a
radical equation
usually involves three main steps.
Isolate the radical on one side of the equation.
Raise each side of the equation to a power equal to the index of the radical to eliminate the radical.
Solve the resulting equation.
Check the results for
extraneous solutions
.
Now we can analyze the given radical equation.
-
2
2
4
x
+
1
3
=
-
1
1
First, let's isolate the radical,
2
4
x
,
on one side of the equation.
-
2
2
4
x
+
1
3
=
-
1
1
SubEqn
LHS
−
1
3
=
RHS
−
1
3
-
2
2
4
x
=
-
2
4
DivEqn
LHS
/
(
-
2
)
=
RHS
/
(
-
2
)
2
4
x
=
1
2
We obtained an isolated radical with index equal to
2
.
Then, we will raise each side of the equation to the power of
2
.
2
4
x
=
1
2
RaiseEqn
LHS
2
=
RHS
2
(
2
4
x
)
2
=
1
2
2
Solve for
x
PowSqrt
(
a
)
2
=
a
2
4
x
=
1
2
2
CalcPow
Calculate power
2
4
x
=
1
4
4
DivEqn
LHS
/
2
4
=
RHS
/
2
4
x
=
6
Next, we will check for extraneous solutions. We do that by substituting
6
for
x
into the original equation. If the substitution produces a true statement, we know that our answer is correct. If it does not, then it is an extraneous solution.
-
2
2
4
x
+
1
3
=
-
1
1
Substitute
x
=
6
-
2
2
4
(
6
)
+
1
3
=
?
-
1
1
Simplify
Multiply
Multiply
-
2
1
4
4
+
1
3
=
?
-
1
1
CalcRoot
Calculate root
-
2
(
1
2
)
+
1
3
=
?
-
1
1
Multiply
Multiply
-
2
4
+
1
3
=
?
-
1
1
AddTerms
Add terms
-
1
1
=
-
1
1
✓
Because our substitution produced a true statement, we know that our answer,
x
=
6
,
is correct.