Download Simplify 243. Simplifying Radicals 11

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
4x2 is a perfect square and a
factor of 40x2.
= 21
4x2 •
Use the Multiplication Property of
Square Roots.
= 21 • 2x
= 42x
10
10
Simplify
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
Divide the numerator and denominator by 3x.
=
25x4
16
Use the Division Property of Square Roots.
=
25 •
16
5x2
= 4
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
3 7
49
3
7
7
7
7
Multiply by
7
7
to make the denominator a
perfect square.
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.
The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
What is the length of the hypotenuse
of this triangle?
a2 + b2 = c2
Use the Pythagorean Theorem.
82 + 152 = c2
Substitute 8 for a and 15 for b.
64 + 225 = c2
289 =
17 = c
c2
Simplify.
Find the principal square root of each side.
Simplify.
The length of the hypotenuse is 17 m.
11-2
The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
A toy fire truck is near a toy building on a table such that the
base of the ladder is 13 cm from the building. The ladder is extended 28
cm to the building. How high above the table is the top of the ladder?
Define: Let b = height (in cm) of the ladder from
a point 9 cm above the table.
Relate: The triangle formed is a right triangle.
Use the Pythagorean Theorem.
11-2
The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
(continued)
Write:
a2 + b2 = c2
132 + b2 = 282
169 + b2 = 784
Substitute.
Simplify.
b2 = 615
Subtract 169 from each side.
b2 =
Find the principal square root of each side.
b
615
24.8
Use a calculator and round to the
nearest tenth.
The height to the top of the ladder is 9 cm higher than 24.8 cm,
so it is about 33.8 cm from the table.
11-2
The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
Determine whether the given lengths are sides of a right
triangle.
a. 5 in., 5 in., and 7 in.
52 + 52
72
25 + 25 49
50 =/ 49
Determine whether a2 + b2 = c2,
where c is the longest side.
Simplify.
This triangle is not a right triangle.
b. 10 cm, 24 cm, and 26 cm
102 + 242 262
Determine whether a2 + b2 = c2,
where c is the longest side.
100 + 576 676
Simplify.
676 = 676
This triangle is a right triangle.
11-2
The Pythagorean Theorem
ALGEBRA 1 LESSON 11-2
If two forces pull at right angles to each other, the
resultant force is represented as the diagonal of a rectangle,
as shown in the diagram. The diagonal forms a right triangle
with two of the perpendicular sides of the rectangle.
For a 50–lb force and a 120–lb force, the resultant force is
130 lb. Are the forces pulling at right angles to each other?
502 + 1202
2500 + 14,400
1302
Determine whether a2 + b2 = c2
where c is the greatest force.
16,900
16,900 = 16,900
The forces of 50 lb and 120 lb are pulling at right angles to each other.
11-2
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).
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
Simplify 8
8
5–
45 = 8
5+
=8
5–
=8
5–3
= (8 – 3)
=5
5
5–
9•5
9•
5
5
5
45.
9 is a perfect square and a factor of 45.
Use the Multiplication Property of
Square Roots.
Simplify 9.
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
51
(1 + 5)
= 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)
5)
5)
Multiply the numerator and the
denominator by the conjugate
of the denominator.
Multiply in the denominator.
Simplify the denominator.
x = – 51(1 – 5)
Divide 102 and –4 by the
common factor –2.
x = 31.51973343
Use a calculator.
2
x 32
The exact width of the painting is – 51(1 – 5) inches.
2
The approximate width of the painting is 32 inches.
11-4
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)
b.
x–5 =4
(
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
Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
Find the domain of each function.
a. y =
x+5
x+5>
–0
Make the radicand >
– 0.
x>
– –5
The domain is the set of all numbers greater than or equal to –5.
b. y = 6
4x – 12
4x – 12 >
–0
Make the radicand >
– 0.
4x >
– 12
x>
– 3
The domain is the set of all numbers greater than or equal to 3.
11-6
Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
The size of a television screen is the length of the screen’s
diagonal d in inches.The equation d = 2A estimates the length of a
diagonal of a television with screen area A.
Graph the function.
Domain
2A >
–0
A>
–0
Screen
Area
(sq. in.)
0
50
100
200
300
400
Length of
Diagonal
(in.)
0
10
14.1
20
24.5
28.3
11-6
Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
Graph y =
y=
x + 4 by translating the graph of
x.
For the graph y =
the graph of y =
11-6
x + 4,
x is shifted 4 units up.
Graphing Square Root Functions
ALGEBRA 1 LESSON 11-6
Graph ƒ(x) =
y=
x + 3 by translating the graph of
x.
For the graph ƒ(x) =
the graph of y =
11-6
x + 3,
x is shifted to the left 3 units.
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
Use the triangle. Find sin A, cos A, and tan A.
opposite leg
6
3
sin A = hypotenuse =
=
10 5
adjacent leg
8
4
cos A = hypotenuse =
=
10 5
opposite leg
tan A = adjacent leg = 6 = 3
8
4
11-7
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
Find sin 40° by using a calculator.
To find sin 40°, press
Use degree mode when finding
trigonometric ratios.
Rounded to the nearest ten-thousandth, the sin 40°
is 0.6428.
11-7
40
.
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
Find the value of x in the triangle.
Step 1: Decide which trigonometric ratio to use.
You know the angle and the length of the hypotenuse.
You are trying to find the adjacent side. Use the cosine.
Step 2: Write an equation and solve.
adjacent leg
cos 30° = hypotenuse
x
cos 30° =
15
x = 15(cos 30°)
15
30
12.99038106
x 13.0
The value of x is about 13.0.
Substitute x for adjacent leg
and 15 for hypotenuse.
Solve for x.
Use a calculator.
Round to the nearest tenth.
11-7
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
Suppose the angle of elevation from a rowboat to the top of a
lighthouse is 708. You know that the lighthouse is 70 ft tall. How far
from the lighthouse is the rowboat? Round your answer to the nearest
foot.
Draw a diagram.
Define: Let x = the distance from the boat to the lighthouse.
Relate: You know the angle of elevation and the opposite side.
You are trying to find the adjacent side. Use the tangent.
11-7
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
(continued)
Write:
opposite leg
tan A = adjacent leg
70
tan 70° = X
x(tan 70°) = 70
Substitute for the angle and the sides.
Multiply each side by x.
x=
70
tan 70
Divide each side by tan 70°.
x
25.4779164
Use a calculator.
x
25
Round to the nearest unit.
The rowboat is about 25 feet from the lighthouse.
11-7
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
A pilot is flying a plane 15,000 ft above the ground. The pilot
begins a 3° descent to an airport runway. How far is the airplane from
the start of the runway (in ground distance)?
Draw a diagram.
Define: Let x = the ground distance from the start of the runway.
Relate: You know the angle of depression and the opposite side.
You are trying to find the adjacent side. Use the tangent.
11-7
Trigonometric Ratios
ALGEBRA 1 LESSON 11-7
(continued)
Write:
opposite leg
tan A = adjacent leg
tan 3° =
15,000
x
x(tan 3°) = 15,000
Substitute for the angle and the sides.
Multiply each side by x.
x=
15,000
tan 3
Divide each side by tan 3°.
x
286217.05
Use a calculator.
x
290,000
Round to the nearest 10,000 feet.
The airplane is about 290,000 feet (or about 55 miles)
from the start of the runway.
11-7