Transcript 7.6 Solving Radical Equations

7.6 Solving Radical
Equations
p.437
Objectives
1. Solve equations that contains radicals
or rational exponents
2. Be able to identify the extraneous
solution(s).
3. Use radical equation to solve some reallife problems.
Ex: Solve for x.
Don’t forget to
check your
solutions!!
x 3
 x  3
2
2 **
x9
** Use this idea to solve the radical equations.
Do
Worksheet 7.6 Practice A #13,14,16,17
Ex: Solve for y.
54 y  2
 4 y  3
Don’t forget to
check your
solutions!!
4
y 3
 y
4
4
 34
y  81
The strategy is
isolating the radical
then taking n-th
power to perform
“Do-Undo”.
Ex: Solve.
Don’t forget to
check your
solutions!!
3
2x  8  4  2
3
2x  8  2
 2x  8   (2)
3
3
2x  8  8
2x  16
x  8
3
Challenge
Worksheet 7.6 Practice A #25, 26
Practice
x5  7
1.
3.
3
9 x 2  0
2
2.
4.
3
6  2 x  2
x2 30
Ex: Solve.
4
3
(a) 3x  243
4
3
(b) 3x  243
4
3
4
3
x  81
x  81
3
4
3
 43 
 x   814
 
 
x
 81 
3
4
x  3
3
x  27
After checking, x = 27 is the solution
x 
4
1
3
 34
x  3
4
x
4

 3 
4 3
3 4
Taking the 4-th root at both sides:
x  33  27
After checking, x = ±27 are the
solutions.
Warning
When you solve the rational exponent
equations, try to AVOID applying
another rational exponent power onto
the existing variable rational exponent
expression. Fail to do so may cause
losing some qualified (valid) answer(s).
Suggestion
Applying the power against to the
denominator in the rational exponent
first, then solve polynomial equation.
Practice
1.
3.
2
3
x  16
2.
4
3
3( x  1)  48
4.
2
3
4 x  100
2
5
5( x  6)  20
Application of Using Radical Equation
P.443 Q. 69
You are trying to determine the
height of a truncated pyramid
that cannot be measured directly.
The height h and slant height l of
a truncated pyramid are related
by the formula
2
D
C
5
A
E
4
B
AC = h = height of the trp. pyramid
BC = l = slant height of the trp. Pyramid = 5
DC = b1 = 2, EB = b2 = 4, AB = ½ (EB - DC) = ½(b2 – b1) = 1
1
BC  l  AC  AB  h  (b2  b1 )2
4
2
2
2
5  h 2  12
5  h 2  12
25  h 2  1
h 2  24
h   24
Note that the answer h   24 is the solution to the
equation
5  h 2  12
However, it is not the solution to this application
question because there is no negative length. So we
discard it and only take
h  24
So, why didn’t one of the
solutions work?




The solution that did not work is
called an extraneous solution.
An extraneous solution is a false
solution.
This is the reason why you MUST
check all solutions!!
The final answer MUST exclude the
extraneous solution.
Ex: Solve.
(a) 12  2x  2 x  0
(b)
12  2 x  2 x
4 x  28  3 2 x
 12  2 x   2 x 
2
4x  28  3 2x  0
2
12  2 x  4 x

4 x  28
  3
2
2x
4 x  28  9  2 x
12  6 x
4 x  28  18 x
2x
28  14 x
2x
Don’t forget to
check your
solutions!!

2
Ex: Solve.
x  4  2x
x  4
2

 2x 
2
x  4x  4  2 x
8  4  2 8
2  4  22
4  16
2  4
44
2 2
x 2  8x  16  2 x
x 2  10x  16  0
x  8x  2  0
x 8  0
x20
x8
x2
Solution
Factor this to solve!
Don’t forget to
check your
solutions!!
Only use
the positive
answers
when
checking
solutions.
Assignment