7.6
Solving Radical Equations
Goals p Solve equations that contain radicals.
p Use radical equations to solve real-life problems.
Your Notes
VOCABULARY
Extraneous solution A solution of a transformed
equation that is not a valid solution of the original
equation
Example 1
Solving a Simple Radical Equation
3
Solve !x" ! 4 " !3.
3
!x" ! 4 " !3
3
Write original equation.
!x" " 1
Isolate radical.
")3 " 13
(!x
Cube each side.
3
x"1
Simplify.
The solution is 1 . Check this in the original equation.
Example 2
Solving an Equation with Rational Exponents
Solve !x2/3 " !49.
!x2/3 " !49
x2/3 " 49
Write original equation.
Isolate radical.
x " (491/2)3
3
Raise each side to ## power.
2
Apply properties of roots.
x " 73 " 343
Simplify.
(x2/3)3/2 " 493/2
The solution is 343 . Check this in the original equation.
168
Algebra 2 Notetaking Guide • Chapter 7
M2ng_nt.07
9/28/04
11:43 AM
Page 169
Your Notes
Example 3
Solving an Equation with One Radical
Solve !3x
"
! 4 " 1 # 7.
Solution
!3x
!4"1#7
"
Write original equation.
!3x
!4# 8
"
Isolate radical.
(!3x
! ")2 # 82
"4
Square each side.
3x ! 4 # 64
Simplify.
3x # 60
Subtract 4 from each side.
x # 20
Divide each side by 3 .
Check x # 20 in the original equation.
!3(
20 "
")
!4"1!7
!64
""1 ! 7
7#7
Substitute for x.
Simplify.
Solution checks .
The solution is 20 .
Example 4
Solving an Equation with Two Radicals
Solve !2x
"
" 5 " !x"
! 3 # 0.
Solution
!2x
" 5 " !x"
"
!3#0
Write original
equation.
!2x
" " # !x
"5
"
!3
Add !x
"
! 3 to
each side.
(!2x
" ")2 # (!x
"5
")
!32
Square each side.
2x " 5 # x ! 3
Simplify.
x#8
Solve for x.
Check x # 8 in the original equation.
!2(
8)"
""
5 " !8
"
!3!0
!11
" " !11
" ! 0
0#0
Substitute for x.
Simplify.
Solution checks .
The solution is 8 .
Lesson 7.6 • Algebra 2 Notetaking Guide
169
M2ng_nt.07
9/28/04
11:43 AM
Page 170
Your Notes
An Equation with an Extraneous Solution
Example 5
Solve x ! 2 " !x".
x # 2 " !x"
Write original equation.
(x # 2)2 " (!x
")2
Square each side.
x2 # 4x ! 4 " x
Expand left side, simplify
right.
x2 # 5x ! 4 " 0
Write in standard form.
(x # 1)(x # 4) " 0
Factor.
(x # 1) " 0 or (x # 4) " 0
Zero product property
x " 1 or x " 4
Simplify.
Check x " 1 and x " 4 in the original equation.
x # 2 " !x"
1 # 2 ! !1
"
#1 $ 1
Write original equation.
Substitute for x.
x # 2 " !x"
4 # 2 ! !4
"
2"2
Simplify.
The only solution is 2 .
Checkpoint Solve the equation.
3
1. !x" ! 7 " 5
#8
3. !5x
"1
# " ! 6 " 13
Homework
10
5. x # 6 " !3x
"
12
170
Algebra 2 Notetaking Guide • Chapter 7
2. 7x1/3 " 14
8
4. !x"
# 2 # !5x
"2
!""0
no solution