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Simplifying Radicals
ALGEBRA 1 LESSON 11-1
(For help, go to Lessons 8-3 and 10-3.)
Complete each equation.
1. a 3 = a2 • a
2. b 7 = b6 • b
3. c 6 = c3 • c
4. d 8 = d4 • d
Find the value of each expression.
5.
4
8.
49
6.
169
11-1
7.
25
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Solutions
1. a3 = a(2 + 1) = a2 • a1
2. b7 = b(6 + 1) = b6 • b1
3. c6 = c(3 + 3) = c3 • c3
4. d8 = d(4 + 4) = d4 • d4
5.
4=2
6.
169 = 13
7.
25 = 5
8.
49 = 7
11-1
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Simplify
243 =
=
= 9
81 • 3
81 •
3
243.
81 is a perfect square and a factor of 243.
3
Use the Multiplication Property of Square Roots.
Simplify
11-1
81.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Simplify
28x7 =
=
4x6 • 7x
4x6 •
= 2x3
7x
28x7.
4x6 is a perfect square and a factor of 28x7.
7x
Use the Multiplication Property of Square Roots.
Simplify
11-1
4x6.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Simplify each radical expression.
a.
12 •
12 •
32
32 =
12 • 32
Use the Multiplication Property of
Square Roots.
=
384
Simplify under the radical.
=
64 • 6
64 is a perfect square and a factor of 384.
=
64 •
= 8
6
6
Use the Multiplication Property of
Square Roots.
Simplify
11-1
64.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
(continued)
b. 7
5x • 3
8x
7
5x • 3
8x = 21
40x2
Multiply the whole numbers and
use the Multiplication Property of
Square Roots.
= 21 4x2 • 10
factor of 40x2.
4x2 is a perfect square and a
= 21 4x2 • 10
Square Roots.
Use the Multiplication Property of
= 21 • 2x
Simplify
= 42x
10
10
Simplify.
11-1
4x2.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Suppose you are looking out a fourth floor window 54 ft above
the ground. Use the formula d = 1.5h to estimate the distance you
can see to the horizon.
d =
1.5h
=
1.5 • 54
Substitute 54 for h.
=
81
Multiply.
=9
Simplify
81.
The distance you can see is 9 miles.
11-1
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Simplify each radical expression.
a.
b.
13
64
13
=
64
13
64
Use the Division Property of Square Roots.
=
13
8
Simplify
49
=
x4
49
x4
Use the Division Property of Square Roots.
64.
49
x4
7
= x2
Simplify
49 and
11-1
x4.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Simplify each radical expression.
a.
120
10
120
=
10
12
Divide.
=
4•3
4 is a perfect square and a factor of 12.
=
4•
=2
3
3
Use the Multiplication Property of Square Roots.
Simplify
11-1
4.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
(continued)
b.
75x5
48x
75x5
=
48x
25x4
16
=
25x4
16
=
=
25 •
16
5x2
4
Divide the numerator and denominator by 3x.
Use the Division Property of Square Roots.
x4
Use the Multiplication Property of
Square Roots.
Simplify
25,
11-1
x4, and
16.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Simplify each radical expression.
a.
3
7
3
=
7
=
=
3
•
7
7
7
Multiply by
7
7
to make the denominator a
perfect square.
3
7
49
3
7
7
Use the Multiplication Property of Square Roots.
Simplify
11-1
49.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
(continued)
Simplify the radical expression.
b.
11
12x3
11
=
12x3
11
•
12x3
3x
3x
=
33x
36x4
Use the Multiplication Property of Square Roots.
=
33x
6x2
Simplify
Multiply by
3x to make the denominator a
3x
perfect square.
11-1
36x4.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Simplify each radical expression.
1.
16 •
8 8
4.
2
a5
2
a
a3
2 2. 4
5.
144
3x
15x3
11-1
48
5
5x
3.
12
36
3
3
The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
(For help, go to Lessons 11-2 and 2-7.)
Find the length of the hypotenuse with the given leg lengths.
If necessary, round to the nearest tenth.
1. a = 3, b = 4
2. a = 2, b = 5
3. a = 3, b = 8
4. a = 7, b = 5
For each set of values, find the mean.
5. x1 = 6, x2 = 14
6. y1 = –4, y2 = 8
7. x1 = –5, x2 = –7
8. y1 = –10, y2 = –3
11-3
The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
Solutions
1. a2 + b2 = c2
2.
32 + 42 = c2
9 + 16 = c2
25 = c2
c = 25 = 5
The length of the hypotenuse is 5.
3. a2 + b2 = c2
4.
32 + 82 = c2
9 + 64 = c2
73 = c2
c = 73 8.5
The length of the hypotenuse is
about 8.5.
6 + 14 20
= 2 = 10
2
6. mean = –4 + 8 = 4 = 2
2
2
5. mean =
a2 + b2 = c2
22 + 52 = c2
4 + 25 = c2
29 = c2
c = 29 5.4
The length of the hypotenuse is about 5.4.
a2 + b2 = c2
72 + 52 = c2
49 + 25 = c2
74 = c2
c = 74 8.6
The length of the hypotenuse is
about 8.6.
–12
–5 + (–7)
=
2 = –6
2
8. mean = –10 + (–3) = –13 = –6.5
2
2
7. mean =
11-3
The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
Find the distance between F(6, –9) and G(9, –4).
d=
( x2 – x1)2 + (y2 – y1)2 Use the distance formula.
d=
(9 – 6)2 + [–4 – (–9)]2 Substitute (9, –4) for (x2, y2)
and (6, –9) for (x1, y1).
d=
32 + 52
Simplify within parentheses.
d=
34
Simplify to find the exact
distance.
d
5.8
Use a calculator. Round to
the nearest tenth.
The distance between F and G is about 5.8 units.
11-3
The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
Find the exact lengths of each side of quadrilateral EFGH.
Then find the perimeter to the nearest tenth.
The perimeter =
EF =
=
=
=
[4 – (–1)]2 + (3 + 5)2
52 + (–2)2
25 + 4
29
FG =
=
=
=
(3 – 4)2 + (–2 – 3)2
(–1)2 + (–5)2
1 + 25
26
GH =
|–2 – 3| = 5
EH =
=
=
=
[–2 – (–1)]2 + (–2 – 5)2
(–1)2 + (–7)2
1 + 49
50
29 +
26 + 5 +
50
11-3
22.6 units.
The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
Find the midpoint of CD.
x1+ x2 y1+ y2
2 ,
2
(–3) + 5 7 + 2
=
, 2
2
2 9
= 2,2
Substitute (–3, 7) for
(x1, y1) and (5, 2) for
(x2, y2).
Simplify each numerator.
1
Write 9 as a mixed
= 1, 42
2
number.
1
The midpoint of CD is M 1, 4 2 .
11-3
The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
A circle is drawn on a coordinate plane. The endpoints
of the diameter are (–3, 5) and (4, –3). What are the coordinates of the
center of the circle?
x1+ x2 y1+ y2
2 ,
2
=
(–3) + 4 5 + (–3)
,
2
2
1 2
= 2,2
1
= 2, 1
1
The center of the circle is at 2 , 1 .
11-3
Substitute (–3, 5) for
(x1, y1) and (4, –3) for
(x2, y2).
The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
1. Find the distance between M(2, –1) and N(–4, 3) to the nearest tenth.
7.2
2. Find the distance between P(–2.5, 3.5) and R(–7.5, 8.5) to the nearest tenth.
7.1
3. Find the midpoint of AB, A(3, 6) and B(0, 2).
(112 , 4)
4. Find the midpoint of CD, C(6, –4) and D(12, –2).
(9, – 3)
5. Find the perimeter of triangle RST to the
nearest tenth of a unit.
9.5 units
11-3
Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
(For help, go to Lesson 11-1.)
Simplify each radical expression.
1.
52
2.
200
3. 4
54
Rationalize each denominator.
5.
3
11
6.
5
8
7.
11-4
15
2x
4.
125x2
Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
Solutions
1.
52 = 4 • 13 = 4 • 13 = 2 13
2.
200 = 100 • 2 = 100 • 2 = 10 2
3. 4 54 = 4 9 • 6 = 4 • 9 • 6 = 4 • 3 • 6 = 12
4.
125x2 =
5.
3 =
11
5 =
8
15 =
2x
6.
7.
25 • 5 • x2 =
3 •
11
5 •
8
15 •
2x
11 =
11
8 =
8
2x =
2x
25 •
5 •
3 • 11 = 33
11 • 11
11
5•8 =
40 =
8
8•8
15 • 2x =
30x
2x
2x • 2x
11-4
x2 = 5x
6
5
4 • 10 =
8
4•
8
10 = 2 10 =
8
10
4
Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
Simplify 4
4
3+
3=4
3+1
= (4 + 1)
=5
3
3+
3
3
3.
Both terms contain
3.
Use the Distributive Property to
combine like radicals.
Simplify.
11-4
Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
5–
Simplify 8
8
5–
45.
45 = 8
5+
9•5
9 is a perfect square and a factor of 45.
=8
5–
9•
=8
5–3
Use the Multiplication Property of
Square Roots.
Simplify 9.
= (8 – 3)
=5
5
5
5
5
Use the Distributive Property to
combine like terms.
Simplify.
11-4
Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
Simplify
5(
8 + 9) =
=
=2
5(
40 + 9
4•
8 + 9).
5
10 + 9
10 + 9
Use the Distributive Property.
5
Use the Multiplication Property
of Square Roots.
5
Simplify.
11-4
Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
Simplify (
(
6–3
6 – 3 21)( 6 + 21)
= 36 + 126 – 3 126 – 3
21)(
441
6+
21).
Use FOIL.
=6–2
126 – 3(21)
Combine like radicals and
simplify 36 and 441.
=6–2
9 • 14 – 63
9 is a perfect square factor of 126.
=6–2
9•
Use the Multiplication Property of
Square Roots.
=6–6
14 – 63
= –57 – 6
14 – 63
Simplify
14
Simplify.
11-4
9.
Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
Simplify
8
7–
=
3
7+
7+
•
3
3
8
7–
3
.
Multiply the numerator and
denominator by the conjugate
of the denominator.
=
8( 7 + 3)
7–3
Multiply in the denominator.
=
8( 7 +
4
Simplify the denominator.
= 2(
7+
=2
7+2
3)
3)
3
Divide 8 and 4 by the common
factor 4.
Simplify the expression.
11-4
Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
A painting has a length : width ratio approximately equal to the
golden ratio (1 + 5 ) : 2. The length of the painting is 51 in. Find the
exact width of the painting in simplest radical form. Then find the
approximate width to the nearest inch.
Define:
51 = length of painting
x = width of painting
Relate: (1 +
Write:
5) : 2 = length : width
(1 + 5) = 51
x
2
x (1 +
5) = 102
x(1 + 5) =
102
(1 + 5)
(1 + 5)
Cross multiply.
Solve for x.
11-4
Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
(continued)
x=
(1 –
102
•
(1 –
(1 + 5)
x=
102(1 – 5)
1–5
x=
102(1 –
–4
5)
x = – 51(1 –
5)
2
x = 31.51973343
5)
5)
Multiply the numerator and the
denominator by the conjugate
of the denominator.
Multiply in the denominator.
Simplify the denominator.
Divide 102 and –4 by the
common factor –2.
Use a calculator.
x 32
The exact width of the painting is – 51(1 –
2
5)
inches.
The approximate width of the painting is 32 inches.
11-4
Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
Simplify each expression.
1. 12 16 – 2
40
4. (
16
3–2
21)(
–123 + 3
7
2.
3+3
20 – 4
–2 5
21)
5
5.
3.
16
5–
–8
11-4
2( 2 + 3
2+3 6
7
5–8
7
3)
Solving Radical Equations
ALGEBRA 1 LESSON 11-5
(For help, go to Lesson 10-3.)
Evaluate each expression for the given value.
x – 3 for x = 16
1.
2.
x + 7 for x = 9 3. 2
x + 3 for x = 1
Simplify each expression.
4.
(
3)2
5.
(
x + 1) 2
11-5
6.
(
2x – 5) 2
Solving Radical Equations
ALGEBRA 1 LESSON 11-5
Solutions
1.
x – 3 for x = 16:
2.
x + 7 for x = 9:
16 – 3 = 4 – 3 = 1
9+7=
3. 2
x + 3 for x = 1: 2
4. (
3)2 = 3
5. (
x + 1)2 = x + 1
6. (
2x – 5)2 = 2x – 5
16 = 4
1+3=2
4=2•2=4
11-5
Solving Radical Equations
ALGEBRA 1 LESSON 11-5
Solve each equation. Check your answers.
a.
x–5 =4
x=9
(
x)2 = 92
Isolate the radical on the left side
of the equation.
Square each side.
x = 81
Check:
x–5
–5
9–5
= 4
4
4
Substitute 81 for x.
4=4
11-5
Solving Radical Equations
ALGEBRA 1 LESSON 11-5
(continued)
x–5 =4
b.
(
x – 5)2 = 42
x–5=9
Square each side.
Solve for x.
x = 21
Check:
x–5
21– 5
16
= 4
= 4
= 4
Substitute 21 for x.
4=4
11-5
Solving Radical Equations
ALGEBRA 1 LESSON 11-5
On a roller coaster ride, your speed in a loop depends on the
height of the hill you have just come down and the radius of the loop in
feet. The equation v = 8 h – 2r gives the velocity v in feet per second
of a car at the top of the loop.
11-5
Solving Radical Equations
ALGEBRA 1 LESSON 11-5
(continued)
The loop on a roller coaster ride has a radius of 18 ft.
Your car has a velocity of 120 ft/s at the top of the loop.
How high is the hill of the loop you have just come
down before going into the loop?
Solve v = 8
120 = 8
120
= 8
8
15 =
h – 2r for h when v = 120 and r = 18.
h – 2(18)
Substitute 120 for v and 18 for r.
h – 2(18)
8
h – 36
(15)2 = ( h – 36)2
225 = h – 36
261 = h
The hill is 261 ft high.
Divide each side by 8 to isolate the radical.
Simplify.
Square both sides.
11-5
Solving Radical Equations
ALGEBRA 1 LESSON 11-5
Solve
(
3x – 4 =
3x – 4)2 = (
2x + 3)2
3x – 4 = 2x + 3
3x = 2x + 7
x=7
Check:
3x – 4 =
3(7) – 4
17 =
2x + 3.
Square both sides.
Simplify.
Add 4 to each side.
Subtract 2x from each side.
2x + 3
2(7) + 3
17
Substitute 7 for x.
The solution is 7.
11-5
Solving Radical Equations
ALGEBRA 1 LESSON 11-5
Solve x =
x + 12.
(x)2 = (
x + 12)2
Square both sides.
x2 = x + 12
x2 – x – 12 = 0
Simplify.
(x – 4)(x + 3) = 0
Solve the quadratic equation by factoring.
(x – 4) = 0 or (x + 3) = 0
x = 4 or
x = –3
Use the Zero–Product Property.
Solve for x.
Check:
x =
x + 12
4
4 + 12
4 = 4
–3
–3 =/
–3 + 12
3
The solution to the original equation is 4.
The value –3 is an extraneous solution.
11-5
Solving Radical Equations
ALGEBRA 1 LESSON 11-5
Solve
3x + 8 = 2.
3x = –6
(
3x)2 = (–6)2
Square both sides.
3x = 36
x = 12
Check:
3x
3(12)
36
6
+8=2
+8 2
+8 2
+ 8 =/ 2
Substitute 12 for x.
x = 12 does not solve the original equation.
3x + 8 = 2 has no solution.
11-5
Solving Radical Equations
ALGEBRA 1 LESSON 11-5
Solve each radical equation.
25
1.
7x – 3 = 4
3.
2x + 7 =
5.
3x + 4 + 5 = 3
2.
7
5x – 8 5
3x – 2 =
4. x =
no solution
11-5
2x + 8
x+2
4
2