Algebra Notes
Chapter 10 – Radical Functions and Geometry
10.1 Square Root Functions
Example 1 Dilation of the Square Root Function
Graph f(x) = 8 x. State the domain and range.
Step 1
Make a table. Choose nonnegative values for x.
x
Step 2
f(x)
0
0
0.5
5.7
1
8
2
11.3
3
13.9
4
16
Plot points and draw a smooth curve.
18
y
16

14
12




10
–5 –4 –3 –2 O 1 2 3 4 5 x
8
The domain is {x | x  0}, and the range is {y | y  0}.
Example 2 Reflection of the Square Root Function
Graph y = -6 x, and compare to the parent graph. State the domain and range.
Make a table of values. Then plot the points on a coordinate system and
y
2
draw a smooth curve that connects them.
O 1 2
-2
x
0
0.5
1
4
9
y
0
-4.2
-6
-12
-18
-4
3 4 5 6 7 8 9 x



-6

-8
Notice that the graph is in the 4th quadrant. It is a dilation
of the graph of y = x that has been reflected across the xaxis. The domain is {x | x  0}, and the range is {y | y  0}.
Example 3 Translation of the Square Root Function
Graph each function, and compare to the parent graph. State the domain and range.
a. g(x) = x – 5
Make a table of values.
x
0
0.5
1
4
9
y
-5
-4.3
-4
-3
-2
f(x)
g(x)
Notice that the values of g(x) are 5 less than those of f(x) =
x. This is a vertical translation 5 units down from the parent
function. The domain is {x | x  0}, and the range is {y | y  –
5}.
b. h(x) = x + 6
g(x)
Make a table of values.
f(x)
x
-6
-2
0
3
y
0
2
2.4
3
This is a horizontal translation 6 units to the left of the parent
function. The domain is {x | x  -6}, and the range is {y | y 
0}.
Real-World Example 4 Analyze a Radical Function
S
6, where S is the surface area
of the cube. Graph the function. Determine the side length of a cube that has a surface area of 112
square inches.
GEOMETRY The side length s of a cube is given by the function s =
Graph the function.
y
7
6
5
To find the side length of the cube, substitute 112 inches for
S.
4
3
2
1
s=
=
S
6
Original function
112
6
S = 112
 4.3 in.
-5
Use a calculator.
The side length of the cube is approximately 4.3 inches.
5
-1
10 15 20 25 30 35
Example 5 Transformations of the Square Root Function
Graph y = 3 x – 2, and compare to the parent graph. State the domain and range.
Make a table of values. Then plot the points on a coordinate
system and draw a smooth curve that connects them.
y
6
5
4
3
2
x
0
1
4
1
9
-2
-1
1
2
3
4
5
6
-1
y
-2
1
4
7
-2
This graph is a dilation of the graph of y = x that has been
translated 2 units down. The domain is {x | x  0}, and the range is {y
| y  –2}.
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10.2 Simplifying Radical Expressions
Example 1 Simplify Square Roots
Simplify 124.
124 = 2  2  31
=
22 
Prime factorization of 124
31
Product Property of Square Roots
= 2 31
Simplify.
Example 2 Multiply Square Roots
Simplify
5  18 .
5  18 =
=
5 
5 2 
= 3 10
2  9
Product Property of Square Roots
32
Product Property of Square Roots
Simplify.
Example 3
Simplify
Simplify a Square Root with Variables
27a 5bc 2 .
27a 5bc 2 = 33 a 5bc 2
=
32  3 
=3
3  a2 
Prime factorization
a4  a  b  c2
Product Property
a  b  c
Simplify.
= 3a2|c| 3ab
Simplify.
Standardized Test Example 4 Rationalizing the Denominator
Which expression is equivalent to
5
?
108
A
5
108
C
B
540
108
D
5
6 3
5 3
18
Read the Test Item
The radical expression needs to be simplified.
Solve the Test Item
5
108
=
5
108

108
108
Multiply by
108
108
.
=
5 108
108
Product Property of Square Roots
=
5 2  2 333
108
Prime factorization
=
30 3
108
5 3
18
The correct choice is D.
=
22 = 2 and
Simplify.
32 = 3
Example 5 Use Conjugates to Rationalize a Denominator
Simplify
5
2 2 3
5
.
2 2+3
=
=
5
2 2 3

2 2 3
The conjugate of 2
2 2 3
5(2 2  3)
(2 2 ) 3
2
2 + 3 is 2 2 – 3.
(a – b)(a + b) = a2 – b2
2
=
10 2  15
89
(2
=
10 2  15
1
Simplify.
= 15 – 10 2
Simplify.
2 )2 = 8
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10.3 Operations with Radical Expressions
Example 1 Add and Subtract Expressions with Like Radicands
Simplify each expression.
a. -3 3 – 3 + 8 3
-3 3 – 3 + 8 3 = (-3 – 1 + 8) 3
=4 3
Distributive Property
Simplify.
b. 7 11 – 4 3 + 6 11 – 2 3
7 11 – 4 3 + 6 11 – 2 3 = (7 + 6) 11 + (-4 – 2) 3
= 13 11 – 6 3
Distributive Property
Simplify.
c. 10 7 + 4 7 – 9 7
10 7 + 4 7 – 9 7 = (10 + 4 – 9) 7
Distributive Property
=5 7
Example 2
Simplify.
Add and Subtract Expressions with Unlike Radicands
Simplify -5 28 + 6 63 – 11 7.
-5 28 + 6 63 – 11 7 = -5 2 2 7 + 6 3 2 7 – 11 7
Prime Factorization
= -5( 22  7) + 6( 32  7) – 11 7
Product Property
= -5(2 7) + 6(3 7) – 11 7
Simplify.
= -10 7 + 18 7 – 11 7
Multiply.
= -3 7
Simplify.
Example 3 Multiply Radical Expressions
Simplify each expression.
a. 4 10  6 5
4 10  6 5 = (4  6)( 10  5)
Associative Property
= (24)( 50)
Multiply.
= (24)(5 2)
Simplify.
= 120 2
Multiply.
b. 5 2(3 3 - 2 2)
5 2(3 3 - 2 2) = (5 2  3 3) – (5 2  2 2)
Distributive Property
= [(5  3)( 2  3)] – [(5  2)( 2  2)]
Associative Property
= [(15)( 6)] – [(10)( 4)]
Multiply.
= (15)( 6) – (10)(2)
Simplify.
= 15 6 – 20
Multiply.
Real-World Example 4 Multiply Radical Expressions
GEOMETRY Find the area of a parallelogram with a base of 3 6 + 5 and a height of 6
– 2. Write in simplest form.
You can find the area of a parallelogram by multiplying the base times the
height and simplifying.
(3 6 + 5)( 6 – 2)
First Terms
A=b•h
Outer Terms
Inner Terms
Last Terms
(3 6)( 6) + (3 6)(– 2) + ( 5)( 6) + ( 5)(– 2)
= 3 36
–
3 12
+
30
–
= 18 – 6 3 + 30 – 10
10
Multiply.
Simplify.
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10.4 Radical Equations
Real-World Example 1 Variable as a Radicand
CLIFF DIVING A diver dives off of a cliff and reaches the water in 3 seconds. The time t in
h
seconds that it takes an object to fall h feet can be approximated by the equation t = 4 . What
was the height of the cliff?
Understand You know the time it took the diver to reach the water and the equation that
relates time and height.
Plan
h
The diver reaches the water in 3 seconds. Replace t with 3 in the equation t = 4
and solve for h.
Solve
h
t= 4
Original equation
h
3= 4
Replace t with 3.
h
3(4) = 4 (4)
12 =
h
Multiply each side by 4.
Simplify.
122 = ( h )2
Square each side of the equation.
144 = h
Simplify.
The height of the cliff was 144 feet.
Check
Check by substituting 144 for h in the original equation. Remember that the time
is 3 seconds.
h
t= 4
?
3
Original equation
144
4
? 12
3 4
t = 3 and h = 144
144 = 12
3=3
Example 2
Solve
Simplify.
Expression as a Radicand
2x -3 – 2 = 7. Check your solution.
2x  3 – 2 = 7
2x  3 = 9
( 2 x  3 )2 = 9 2
2x – 3 = 81
2x = 84
x = 42
Original equation
Add 2 to each side.
Square each side.
Simplify.
Add 3 to each side.
Divide each side by 2.
Example 3 Variable on Each Side
Solve
x + 5 – 3 = x.
x5 –3=x
Original equation
x5 =x+3
Add 3 to each side.
( x  5 )2 = (x + 3)2
Square each side.
x + 5 = x2 + 6x + 9
Simplify.
0 = x2 + 5x + 4
Subtract x and 5 from each side.
0 = (x + 1)(x + 4)
Factor.
x + 1 = 0 or x + 4 = 0
x = –1
Zero Product Property
x = –4
Solve.
CHECK
x5 –3=x
Original equation
x5 –3=x
Substitution
 4  5 – 3  –4
?
?
 1 5 – 3  –1
?
?
4 – 3  –1
Simplify.
1 – 3  –4
Simplify.
1 – 3  –4
?
?
2 – 3  –1
–1 = –1

–2  –4

Since –4 does not satisfy the original equation, -1 is the only solution.
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10.5 The Pythagorean Theorem
Example 1 Find the Length of a Side
Find the length of each missing side. If necessary, round to the nearest hundredth.
a. c 2 = a 2 + b 2
Pythagorean Theorem
c 2 = 12 2 + 9 2
a = 12 and b = 9
c
c 2 = 144 + 81
Evaluate squares.
12
c 2 = 225
Simplify.
c= 
225
c =  15
9
Take the square root of each side.
(±15)2 = 225
The length of a side cannot be negative, so the length of the missing side is 15 units.
b.
c2 = a2 + b2
Pythagorean Theorem
37 2 = a 2 + 19 2
c = 37 and b = 19
1369 = a 2 + 361
Evaluate squares.
1008 = a 2
Subtract 361 from each side.
 1008 = a
 31.75 = a
31.75  a
Take the square root of each side.
Use a calculator to evaluate
Use the positive value.
The length of the missing side is 31.75 units.
1008 .
Real-World Example 2 Find the Length of a Side
LADDERS A ladder 14 feet long is leaning up against
a house as shown at the right. How far from the house is
the ladder placed?
142 = 122 + d2
Pythagorean Theorem
196 = 144 + d2
Evaluate squares.
14 ft
12 ft
d
52 = d2 Subtract 144 from each side.
 52 = d Take the square root of each side.
±7.2  d
7.2  d
Use a calculator to evaluate
52 .
Use the positive value.
The ladder is approximately 7.2 feet from the house.
Example 3 Check for Right Triangles
Determine whether 14, 15, and 19 can be sides of a right triangle.
Since the measure of the longest side is 19, let c = 19, a = 14, and b = 15.
Then determine whether c2 = a2 + b2.
c2 = a2 + b2
Pythagorean Theorem
19 2
14 2 + 15 2
c = 19, a = 14, and b = 15
361
196 + 225
Evaluate squares.
361  421
Add.
Since c2 ≠ a2 + b2, segments with these measures cannot form a right triangle.
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10.6 Trigonometric Ratios
Example 1 Find Sine, Cosine, and Tangent Ratios
Find the values of the three trigonometric ratios for angle A.
Step 1
Use the Pythagorean Theorem to find BC.
a2 + b2 = c2
Step 2
B
26
Pythagorean Theorem
a2 + 242 = 262
b = 24 and c = 26
a + 576 = 676
Simplify.
a2 = 100
Subtract 576 from each side.
a = 10
Take the square root of each side.
A
24
Use the side lengths to write the trigonometric ratios.
sin A =
opp 10 5


hyp 26 13
cos A =
adj 24 12


hyp 26 13
Example 2 Use a Calculator to Evaluate Expressions
Use a calculator to find sin 75 to the nearest ten-thousandth.
KEYSTROKES: SIN 75 )
ENTER 0.9659258263
Round to the nearest ten-thousandth, sin 75°  0.9659.
tan A =
opp 10 5


adj 24 12
a
C
Example 3 Solve a Triangle
21
M
N
Solve the right triangle. Round each side length to the nearest tenth.
Step 1
Find the measure of  N. The sum of the
measures of the angles in a triangle is 180.
35
L
180° – (90° + 35°) = 55°
The measure of  N is 55°.
Step 2
Find the measure of LM . Since you are given the measure of the side opposite L and
are finding the measure of the side adjacent to L, use the tangent ratio.
21
tan 35° = n
n tan 35° = 21
n=
21
tan 35
n  30.0
Definition of tangent
Multiply each side by n.
Divide each side by tan 35.
Use a calculator.
So, the measure of LM is about 30.0.
Step 3
Find the measure of LN . Since you are given the measure of the side opposite L and
are finding the measure of the hypotenuse, use the sine ratio.
21
sin 35° = m
m sin 35° = 21
m=
21
sin 35
m  36.6
Definition of sine
Multiply each side by m.
Divide each side by sin 35.
Use a calculator.
So, the measure of LN is about 36.6.
Real-World Example 4 Find a Missing Side Length
DECORATING A plant shelf 8 inches wide makes a 56 angle with
the brace b. Approximately how many inches long is the brace?
8 in.
56
You need to find the hypotenuse. You are given the side adjacent to the
56° angle. Use the cosine ratio.
8
cos 56° = b
Definition of cosine
b  cos 56° = 8
b=
b
Multiply each side by b.
8
cos 56
b  14.3
Divide each side by cos 56.
Use a calculator.
So, the brace is about 14.3 inches long.
S
Example 5 Find a Missing Angle Measure
Find mT to the nearest degree.
T
You know the measure of the side opposite to T and the
measure of the hypotenuse. Use the sine ratio.
15
15
sin T = 23
R
Definition of sine
23
Use a calculator and the [COS-1] function to find the measure
of the angle.
KEYSTROKES:
2ND [SIN-1] 15

23 )
ENTER 40.70570683
So, mT  40.
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