Download 10 pts possible Evaluate the expression. 1. 2. (x – 5)2 Solve the

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
Bell Quiz 6-6
Evaluate the expression.
64
3 pts
1.
3 pts
2. (x – 5)2
Solve the equation.
4 pts
3. x2 + 6x + 9 = x + 45
10 pts
possible
Section 6-6
1
6-6
Chapter 6: Rational Exponents and Radical Functions
Section 6-6
2
6-6
Chapter 6: Rational Exponents and Radical Functions
Section 6-6
3
Definition
Radical Equations
Equations with radicals that have variables in their
radicands are called radical equations
Example:
Section 6-6
4
Key Concept
Solving Radical Equations
To solve a radical equation, follow these steps:
Step 1: Isolate the radical on one side of the equation (if necessary).
Step 2: Raise each side of the equation to the same power to
eliminate the radical.
Step 3: Solve the equation.
Section 6-6
5
EXAMPLE 1
Solve a radical equation
3
Solve √
2x+7 = 3.
Section 6-6
6
EXAMPLE 1
Solve a radical equation
3
Solve √
2x+7 = 3.
3
√ 2x+7 = 3
( √3 2x+7 )3 = 33
2x+7 = 27
2x = 20
x = 10
Write original equation.
Cube each side to eliminate the radical.
Simplify.
Subtract 7 from each side.
Divide each side by 2.
Section 6-6
7
EXAMPLE 1
Solve a radical equation
CHECK
Check x = 10 in the original equation.
3
√ 2(10)+7 =? 3
3
√ 27 =? 3
3 = 3
Substitute 10 for x.
Simplify.
Solution checks.
Section 6-6
8
GUIDED PRACTICE
for Example 1
Solve equation. Check your solution.
1.
3√
x – 9 = –1
x = 512
2. (√ x+25 ) = 4
x = –9
Section 6-6
3.
(23√ x – 3 ) = 4
x = 11
9
EXAMPLE 2
Solve a radical equation given a function
Section 6-6
10
Rational Exponents
When an equation contains a power with a rational exponent
a.k.a fractional exponent
Step 1: Isolate the power.
Step 2: Raise each side of the equation to the reciprocal of the
exponent
Step 3: Solve the equation.
Section 6-6
11
EXAMPLE 3
Standardized Test Practice
Section 6-6
12
EXAMPLE 3
Standardized Test Practice
SOLUTION
4x2/3 = 36
x2/3 = 9
(x2/3)3/2 = 93/2
x = 27
Write original equation.
Divide each side by 4.
3
Raise each side to the power 2 .
Simplify.
ANSWER
The correct answer is D.
Section 6-6
13
EXAMPLE 4
Solve an equation with a rational exponent
Solve (x + 2)3/4 – 1 = 7.
Section 6-6
14
EXAMPLE 4
Solve an equation with a rational exponent
Solve (x + 2)3/4 – 1 = 7.
(x + 2)3/4 – 1 = 7
(x + 2)3/4 = 8
(x +
2)3/4
4/3
= 8 4/3
Write original equation.
Add 1 to each side.
Raise each side to the power 4 .
3
x + 2 = (8 1/3)4
Apply properties of exponents.
x + 2 = 24
Simplify.
x + 2 = 16
Simplify.
x = 14
Subtract 2 from each side.
Section 6-6
15
for Examples 3 and 4
GUIDED PRACTICE
Solve the equation.
4. 3x3/2 = 375
5.
6.
–2x3/4 = –16
x = 25
x = 16
– 2 x1/5 = –2 x = 243
3
Section 6-6
16
GUIDED PRACTICE
for Examples 3 and 4
Solve the equation.
7.
(x + 3)5/2 = 32
8. (x – 5)5/3 = 243
9.
(x + 2)2/3 +3 = 7
x=1
x = 32
x = –10 or 6
Section 6-6
17
Extraneous Solutions
Raising each side of an equation to the same power may introduce
extraneous solutions.
a.k.a It will create extra answers that are not answers to the original
problem.
Always check solutions in original equation.
Section 6-6
18
EXAMPLE 5
Solve an equation with an extraneous solution
Solve x + 1 = √7x + 15 . Check for extraneous solutions.
Section 6-6
19
EXAMPLE 5
Solve an equation with an extraneous solution
Solve x + 1 = √7x + 15
x + 1 = √ 7x + 15
Write original equation.
(x + 1)2 = (√7x + 15)2
Square each side.
Expand left side and simplify
right side.
x2 + 2x + 1 = 7x + 15
x2 – 5x – 14 = 0
Write in standard form.
Factor.
(x – 7)(x + 2) = 0
x – 7 = 0 or x + 2 = 0
x = 7 or
x = –2
Zero-product property
Solve for x.
Section 6-6
20
EXAMPLE 5
Solve an equation with an extraneous solution
CHECK
Check x = 7 in the
original equation.
Check x = –2 in the
original equation.
x + 1 = √7x + 15
x + 1 = √7x + 15
?
?
–2 + 1 = √ 7(–2) + 15
7 + 1 = √ 7(7) + 15
8 = √ 64
?
–1 = √ 1
8=8
–1 =/ 1
ANSWER
?
The only solution is 7.
(The apparent solution –2 is extraneous.)
Section 6-6
21
GUIDED PRACTICE
for Examples 5
Solve the equation. Check for extraneous solutions
10.
−
=
11.
x = 1, ¼
10x + 9 = x + 3
x = 0,4
Section 6-6
22
HOMEWORK
Sec 6-6 (pg 456)
3-21 every 3rd,
23-25, 34-38
Section 6-6
23
Related documents