6.6 Solving Radical Equations
Algebra 2
What is a Radical Equation?
• A Radical Equation is
an equation that has
a variable in a
radicand or has a
variable with a
rational exponent.
3  x  10
2
3
yes
( x  2)  25
yes
3  x  10
no
EXAMPLE 1 – Solving a Radical
Equation
5x  1  6  0
5x  1  6
2
2
5 x  1 ( 6 )
Square both sides to get rid of
the square root
5x  1  36
5x  35
x7
Check your solutions for
extraneous solutions!
EXAMPLE 2
x  15  3  x
2
16  15  3  16
2
x  15 (3  x )
NOTE: when checking solutions only use
positive result of radical. +4 in this case
x  15  (3  x )(3  x)
1 3 4
1  1
x  15  9  6 x  x
NO SOLUTION
 15  9  6 x
 24  6 x
4 x
16  x
Let’s Double Check
that this works
Note: You will get
Extraneous
Solutions from time
to time – always do
a quick check
Let’s Try Some
2  3x  2  6
2
3
2( x  2)  50
Let’s Try Some
Ex. 3
Ex. 4
2  3x  2  6
2
3
2( x  2)  50
Ex. 5 SOLVING MORE
COMPLEX EQUATIONS
4( x  1)  101  20
2
( x  1)   9i 2
4( x  1)  81
2
( x  1)  814
( x  1)  814
2
+ because we
are taking an
even power
(square root)
of both sides
( x  1)  i
81
x  1  9i 2
1  9i 2
4
( x  1)  i 9 2
1  9i 2
6-6A Assignment
RADICAL EQUATIONS
3(5n  1)  2  0
1
3
3(5n  1)  2
1
(5n  1) 3  2 3
1 3
3
(5n  1) 3  2 3 
(5n  1)  8 27 
1
3
5n  8 27   1
5n  8 27   27 27
5n  35 27 
ISOLATE RADICAL / RATIONAL
RAISE BOTH SIDES
TO RECIPROCAL POWER
SOLVE FOR THE VARIABLE
n  7 27 
SOLVING MORE COMPLEX
EQUATIONS
(2 x  1)
0.5
 (3x  4)
0.25
0
(2 x  1)  (3x  4)
0.5 4
0.25 4
[( 2 x  1) ]  [(3x  4) ]
2
(2 x  1)  3x  4
0.5
0.25
Raise each side to the 4th
power. This will get you
integer powers – much
easier to work with!
4 x 2  4 x  1  3x  4
4x2  x  3  0
3
x  ,1
4
Factor
Let’s Try Some . . . check for
extraneous solutions
(2 x  1)
0.5
 (3x  4)
0.25
0
Let’s Try Some . . . check for
extraneous solutions
(2 x  1)
0.5
 (3x  4)
0.25
0
3
x  ,1
4