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Solving Radical Equations
Choose Course
Algebra 2
Radical Functions
Solving Radical Equations
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Solving Radical Equations 1.14 - Solution
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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.
5
+
4
x
−
5
=
1
2
First, let's isolate the radical,
4
x
−
5
,
on one side of the equation.
5
+
4
x
−
5
=
1
2
SubEqn
LHS
−
5
=
RHS
−
5
4
x
−
5
=
7
We get an isolated radical with index equal to
2
.
Then, we will raise each side of the equation to the power of
2
.
4
x
−
5
=
7
RaiseEqn
LHS
2
=
RHS
2
(
4
x
−
5
)
2
=
7
2
Solve for
x
PowSqrt
(
a
)
2
=
a
4
x
−
5
=
7
2
CalcPow
Calculate power
4
x
−
5
=
4
9
AddEqn
LHS
+
5
=
RHS
+
5
4
x
=
5
4
DivEqn
LHS
/
4
=
RHS
/
4
x
=
4
5
4
ReduceFrac
b
a
=
b
/
2
a
/
2
x
=
2
2
7
Next, we will check for extraneous solutions. We do that by substituting
2
2
7
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.
5
+
4
x
−
5
=
1
2
Substitute
x
=
2
2
7
5
+
4
(
2
2
7
)
−
5
=
?
1
2
Simplify
Multiply
Multiply
5
+
2
1
0
8
−
5
=
?
1
2
CalcQuot
Calculate quotient
5
+
5
4
−
5
=
?
1
2
SubTerms
Subtract terms
5
+
4
9
=
?
1
2
CalcRoot
Calculate root
5
+
7
=
?
1
2
AddTerms
Add terms
1
2
=
1
2
✓
Because our substitution produced a true statement we know that our answer,
x
=
2
2
7
,
is correct.