Transcript 7-5 Solving Square Root and Other Radical Equations

7-5 Solving Square
Root and Other Radical
Equations
Objectives
Solving Radical Equations
Vocabulary
If
n
n
n
n
,
then
( x )  k and x  k n
x k
Solving Radical Equations
Solve –10 +
–10 +
(
2x + 1 = –5.
2x + 1 = –5
2x + 1 = 5
Isolate the radical.
2x + 1 )2 = 52
2x + 1 = 25
2x = 24
x = 12
Square both sides.
Check: –10 + 2x + 1 = –5
–10 + 2(12) + 1 –5
–10 + 25 –5
–10 + 5 –5
–5 = –5
Solving Radical Equations with
Rational Exponents
3
5
Solve 3(x + 1) = 24.
3
5
3(x + 1) = 24
3
5
(x + 1) = 8
3
5
5
3
5
3
1)1
5
3
((x + 1) ) = 8
(x +
Divide each side by 3.
=8
5
Raise both sides to the 3 power.
3
5
Multiply the exponents 5 and 3 .
x + 1 = 32
x = 31
Simplify.
3
Check: 3(x + 1) 5 = 24
3
3(31 + 1) 5 24
3
5
3(2 )5 24
3(2)3 24
24 = 24
Real World Example
An artist wants to make a plastic sphere for a sculpture. The
plastic weighs 0.8 ounce per cubic inch. The maximum weight of the
sphere is to be 80 pounds. The formula for the volume V of a sphere
4
is V = • r 3, where r is the radius of the sphere.
3
What is the maximum radius the sphere can have?
Relate: volume of sphere • density of plastic <
– maximum weight
Define: Let r = radius in inches.
4
Write: 3 •
r 3 • 0.8 <
– 80
Continued
(continued)
r3
4
3
• 0.8 <
– 80
3 • 80
r3 <
– 4 • • 0.8
75
r3 <
–
r <
– 2.88
Use a calculator.
The maximum radius is about 2.88 inches.
Checking for Extraneous
Solutions
Solve
x + 2 – 3 = 2x. Check for extraneous solutions.
x + 2 – 3 = 2x
(
x + 2 = 2x + 3
Isolate the radical.
x + 2)2 = (2x + 3)2
Square both sides.
x + 2 = 4x2 + 12x + 9
Simplify.
0 = 4x2 + 11x + 7
Combine like terms.
0 = (x + 1)(4x + 7)
Factor.
x+1=0
or 4x + 7 = 0
x = –1 or
Factor Theorem
7
x=–4
Continued
(continued)
Check:
x + 2 – 3 = 2x
–1 + 2 – 3
2(–1)
x + 2 – 3 = 2x
–7 +2–3
4
–2
1 –3
4
–2 = –2
1 –3
2
1–3
2 –7
4
–7
2
–7
2
– 5 =/ – 7
2
2
The only solution is –1.
Solving Equations with Two
Rational Exponents
2
3
1
3
Solve (x + 1) – (9x + 1) = 0. Check for extraneous
solutions.
2
3
1
3
(x + 1) – (9x + 1) = 0
2
3
(x + 1) = (9x + 1)
2
3 3
1
3
1
3 3
((x + 1) ) = ((9x + 1) )
(x + 1)2 = 9x + 1
x2 + 2x + 1 = 9x + 1
x2 – 7x = 0
x(x – 7) = 0
x = 0 or x = 7
Continued
(continued)
Check:
1
2
(x + 1) 3 – (9x + 1) 3 = 0
2
3
(0 + 1) – (9(0) + 1)
2
3
(1) – (1)
2
3
1
3
1
3
1
(1) –
(12) 3
2
3
2
3
1 –1
Both 0 and 7 are solutions.
0
0
0
=0
2
1
(x + 1) 3 – (9x + 1) 3 = 0
2
3
(7 + 1) – (9(7) + 1)
1
3
1
3
2
3
(8) – (64 )
2
3
(8) –
1
(82) 3
2
3
2
3
0
0
0
8 –8 =0
Homework
p 394 # 1, 2, 7, 8, 15, 16