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Solving Radical Equations
Choose Course
Algebra 2
Radical Functions
Solving Radical Equations
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Solving Radical Equations 1.12 - Solution
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Return to Solving Radical Equations
a
We try a solution by substituting the value into the original equation,
x
=
4
,
and investigate if the left-hand side and the right-hand side are equal. We start with
x
=
1
6
.
x
=
4
Substitute
x
=
1
6
1
6
=
?
4
CalcRoot
Calculate root
4
=
4
x
=
1
6
solves the equation. Let's now test
x
=
2
5
.
x
=
4
Substitute
x
=
2
5
2
5
=
?
4
CalcRoot
Calculate root
5
=
4
Since the left- and right-hand side do not become equal,
x
=
2
5
is not a solution to the equation.
b
We use the same method as in Part A when we now test the values in the equation
x
+
2
0
=
9
x
.
x
+
2
0
=
9
x
Substitute
x
=
1
6
1
6
+
2
0
=
?
9
1
6
AddTerms
Add terms
3
6
=
?
9
⋅
1
6
CalcRoot
Calculate root
3
6
=
?
9
⋅
4
Multiply
Multiply
3
6
=
3
6
x
=
1
6
solves the equation. Next we need to test
x
=
2
5
.
x
+
2
0
=
9
x
Substitute
x
=
2
5
2
5
+
2
0
=
?
9
2
5
AddTerms
Add terms
4
5
=
?
9
⋅
2
5
CalcRoot
Calculate root
4
5
=
?
9
⋅
5
Multiply
Multiply
4
5
=
4
5
Both
x
=
1
6
and
x
=
2
5
are solutions to the radical equation
x
+
2
0
=
9
x
.
c
Finally, we will repeat the process with the last equation.
x
−
9
=
x
−
2
1
Substitute
x
=
1
6
1
6
−
9
=
?
1
6
−
2
1
SubTerm
Subtract term
7
=
?
-
5
UseCalc
Use a calculator
2
.
6
4
5
7
5
…
=
-
5
Here
x
=
1
6
is
not
a solution. Can
x
=
2
5
be one?
x
−
9
=
x
−
2
1
Substitute
x
=
2
5
2
5
−
9
=
?
2
5
−
2
1
SubTerm
Subtract term
1
6
=
?
4
CalcRoot
Calculate root
4
=
4
Yes, it can!
