1. Complete each equation. 1. a 3 = a2 • a 2. b 7 = b6 • b
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.
11-1

2. 1. a3 = a(2 + 1) = a2 • a1 2. b7 = b(6 + 1) = b6 • b1
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
= = 13
= = 7
11-1

3. 243 = 81 • 3 81 is a perfect square and a factor of 243.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Simplify
243 = • is a perfect square and a factor of 243.
= • Use the Multiplication Property of Square Roots.
= Simplify
11-1

4. 28x7 = 4x6 • 7x 4x6 is a perfect square and a factor of 28x7.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Simplify x7.
28x7 = 4x6 • 7x 4x6 is a perfect square and a factor of 28x7.
= 4x6 • 7x Use the Multiplication Property of Square Roots.
= 2x3 7x Simplify 4x6.
11-1

5. Simplify each radical expression.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Simplify each radical expression.
a •
12 • = • 32 Use the Multiplication Property of
Square Roots.
= Simplify under the radical.
= • 6 64 is a perfect square and a factor of 384.
= • 6 Use the Multiplication Property of
Square Roots.
= Simplify
11-1

6. 7 5x • 3 8x = 21 40x2 Multiply the whole numbers and
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
(continued)
b x • x
7 5x • x = x2 Multiply the whole numbers and
use the Multiplication Property of
Square Roots.
= x2 • 10 4x2 is a perfect square and a
factor of 40x2.
= x2 • Use the Multiplication Property of
Square Roots.
= 21 • 2x Simplify 4x2.
= 42x Simplify.
11-1

7. The distance you can see is 9 miles.
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 = h to estimate the distance you can see to the horizon.
d = h
= • 54 Substitute 54 for h.
= Multiply.
= 9 Simplify
The distance you can see is 9 miles.
11-1

8. Simplify each radical expression.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Simplify each radical expression. a. 13 64 b. 49 x4
= Use the Division Property of Square Roots. 13 64
= Simplify 13 8
= Use the Division Property of Square Roots. 49 x4 7 x2
= Simplify and x4.
11-1

9. Simplify each radical expression.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Simplify each radical expression. a.
120 10
= Divide.
120 10
= 4 • 3 4 is a perfect square and a factor of 12.
= 4 • 3 Use the Multiplication Property of Square Roots.
= Simplify
11-1

10. = Divide the numerator and denominator by 3x.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
(continued) b.
75x5
48x
= Divide the numerator and denominator by 3x.
75x5
48x
25x4 16
= Use the Division Property of Square Roots.
25x4 16
= Use the Multiplication Property of
Square Roots.
25 • x4 16
= Simplify , x4, and
5x2 4
11-1

11. Simplify each radical expression.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Simplify each radical expression. a. 3 7 3 7
= • Multiply by to make the denominator a
perfect square.
= Use the Multiplication Property of Square Roots.
3 7 49
= Simplify
3 7 7
11-1

12. Simplify the radical expression.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
(continued)
Simplify the radical expression. b. 11
12x3
= • Multiply by to make the denominator a
perfect square. 3x 11
12x3
= Use the Multiplication Property of Square Roots.
33x
36x4
= Simplify x4.
33x
6x2
11-1

13. Simplify each radical expression.
Simplifying Radicals
ALGEBRA 1 LESSON 11-1
Simplify each radical expression. • 12 36 3
8 2 48 2 a5
2 a a3 3x
15x3 5 5x
11-1

14. 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 = y1 = –4, y2 = 8
7. x1 = –5, x2 = –7 8. y1 = –10, y2 = –3
11-3

15. The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
Solutions
1. a2 + b2 = c2 2. a2 + b2 = c2
= c = c2
= c = c2
25 = c = c2
c = = c =
The length of the hypotenuse is 5. The length of the hypotenuse is about 5.4.
3. a2 + b2 = c a2 + b2 = c2
= c = c2
= c = c2
73 = c = c2
c = c =
The length of the hypotenuse is The length of the hypotenuse is
about about 8.6.
5. mean = = = 10
6. mean = = = 2
6 + 14 2 20
–4 + 8 4
7. mean = = = –6
8. mean = = = –6.5
–5 + (–7)
–10 + (–3)
–12
–13
11-3

16. 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 = Simplify within parentheses.
d = Simplify to find the exact
distance.
d Use a calculator. Round to
the nearest tenth.
The distance between F and G is about 5.8 units.
11-3

17. 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.
EF = [4 – (–1)]2 + (3 + 5)2
= 52 + (–2)2 =
= 29
FG = (3 – 4)2 + (–2 – 3)2
= (–1)2 + (–5)2 =
= 26
GH = |–2 – 3| = 5
EH = [–2 – (–1)]2 + (–2 – 5)2
= (–1)2 + (–7)2 =
= 50
The perimeter = units.
11-3

18. The Distance and Midpoint Formulas
ALGEBRA 1 LESSON 11-3
Find the midpoint of CD.
, = , Substitute (–3, 7) for
(x1, y1) and (5, 2) for
(x2, y2).
7 + 2 2
x1+ x2
y1+ y2
(–3) + 5
= , Simplify each numerator. 2 9
= 1, 4 Write as a mixed
number. 9 2 1
The midpoint of CD is M 1, 1 2
11-3

19. 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?
, = , Substitute (–3, 5) for
(x1, y1) and (4, –3) for
(x2, y2).
5 + (–3) 2
x1+ x2
y1+ y2
(–3) + 4
= , = , 1 1 2
The center of the circle is at , 1 . 1 2
11-3

20. 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.
2. Find the distance between P(–2.5, 3.5) and R(–7.5, 8.5) to the nearest tenth.
3. Find the midpoint of AB, A(3, 6) and B(0, 2).
4. Find the midpoint of CD, C(6, –4) and D(12, –2).
5. Find the perimeter of triangle RST to the
nearest tenth of a unit.
7.2
7.1
(1 , 4) 1 2
(9, – 3)
9.5 units
11-3

21. Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
(For help, go to Lesson 11-1.)
Simplify each radical expression. x2
Rationalize each denominator. 3 11 5 8 15 2x
11-4

22. Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
Solutions
5. = • = =
6. = • = = = = = =
7. = • = = 3 11
3 •
11 • 5 8
5 • 8
8 • 8
4 • 10
4 • 10 40
2 10 10 4 15 2x
15 • 2x
2x • 2x
30x
= • 13 = • =
= • 2 = • 2 =
= • 6 = 4 • • = 4 • 3 • 6 =
x2 = • 5 • x2 = • • x2 = 5x 5
11-4

23. Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
Simplify
= Both terms contain
= (4 + 1) 3 Use the Distributive Property to
combine like radicals.
= Simplify.
11-4

24. Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
Simplify –
– = • 5 9 is a perfect square and a factor of 45.
= – 9 • 5 Use the Multiplication Property of
Square Roots.
= – Simplify
= (8 – 3) 5 Use the Distributive Property to
combine like terms.
= Simplify.
11-4

25. Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
Simplify 5( ).
5( ) = Use the Distributive Property.
= 4 • Use the Multiplication Property
of Square Roots.
= Simplify.
11-4

26. Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
Simplify ( 6 – )( ).
( 6 – )( )
= – – Use FOIL.
= 6 – – 3(21) Combine like radicals and
simplify and
= 6 – • 14 – 63 9 is a perfect square factor of 126.
= 6 – • – 63 Use the Multiplication Property of
Square Roots.
= 6 – – 63 Simplify
= –57 – Simplify.
11-4

27. Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
Simplify . 8
7 – 3
= • Multiply the numerator and
denominator by the conjugate
of the denominator. 8
7 – 3
= Multiply in the denominator.
8( )
7 – 3
= Simplify the denominator.
8( ) 4
= 2( ) Divide 8 and 4 by the common
factor 4.
= Simplify the expression.
11-4

28. Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
A painting has a length : width ratio approximately equal to the golden ratio ( ) : 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: ( ) : 2 = length : width
Write: =
x ( ) = 102 Cross multiply.
= Solve for x.
102
( )
x( ) 51 x 2
11-4

29. Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4
(continued)
x = • Multiply the numerator and the
denominator by the conjugate
of the denominator.
(1 – 5)
102
( )
x = Multiply in the denominator.
102(1 – 5)
1 – 5
x = Simplify the denominator.
102(1 – 5) –4
x = Divide 102 and –4 by the
common factor –2.
– 51(1 – 5) 2
x = Use a calculator.
x 32
The exact width of the painting is inches.
The approximate width of the painting is 32 inches.
– 51(1 – 5) 2
11-4

30. Operations with Radical Expressions
ALGEBRA 1 LESSON 11-4 16
5 – 7
Simplify each expression.
– – ( )
4. ( 3 – )( ) 5. 40
–2 5 –
– –
11-4

31. 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 = x + 7 for x = x + 3 for x = 1
Simplify each expression.
4. ( 3)2 5. ( x + 1) ( 2x – 5) 2
11-5

32. Solving Radical Equations
ALGEBRA 1 LESSON 11-5
Solutions
x – 3 for x = 16: – 3 = 4 – 3 = 1
x + 7 for x = 9: = = 4
x + 3 for x = 1: = = 2 • 2 = 4
4. ( 3)2 = 3
5. ( x + 1)2 = x + 1
6. ( 2x – 5)2 = 2x – 5
11-5

33. Solving Radical Equations
ALGEBRA 1 LESSON 11-5
Solve each equation. Check your answers.
a x – 5 = 4
x = 9 Isolate the radical on the left side
of the equation.
( x)2 = 92 Square each side.
x = 81
Check: x – 5 = 4
– Substitute 81 for x.
9 –
4 = 4
11-5

34. Solving Radical Equations
ALGEBRA 1 LESSON 11-5
(continued)
b x – 5 = 4
( x – 5)2 = 42 Square each side.
x – 5 = 9 Solve for x.
x = 21
Check: x – 5 = 4
21– 5 = 4 Substitute 21 for x.
16 = 4
4 = 4
11-5

35. 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

36. 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 h – 2r for h when v = 120 and r = 18.
120 = 8 h – 2(18) Substitute 120 for v and 18 for r.
= Divide each side by 8 to isolate the radical.
15 = h – 36 Simplify.
8 h – 2(18) 8
120
(15)2 = ( h – 36)2 Square both sides.
225 = h – 36
261 = h
The hill is 261 ft high.
11-5

37. Solving Radical Equations
ALGEBRA 1 LESSON 11-5
Solve 3x – 4 = 2x + 3.
( 3x – 4)2 = ( 2x + 3)2 Square both sides.
3x – 4 = 2x + 3 Simplify.
3x = 2x + 7 Add 4 to each side.
x = 7 Subtract 2x from each side.
Check: x – 4 = 2x + 3
3(7) – (7) + 3 Substitute 7 for x.
17 = 17
The solution is 7.
11-5

38. 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 Use the Zero–Product Property.
x = 4  or   x = –3 Solve for x.
Check: x = x + 12
– –3 + 12
4 = 4 –3 = 3 /
The solution to the original equation is 4.
The value –3 is an extraneous solution.
11-5

39. Solving Radical Equations
ALGEBRA 1 LESSON 11-5
Solve 3x + 8 = 2.
3x = –6
( 3x)2 = (–6) Square both sides.
3x = 36
x = 12
Check: x + 8 = 2
3(12)    Substitute 12 for x.
= 2   x = 12 does not solve the original equation. /
3x + 8 = 2 has no solution.
11-5

40. Solving Radical Equations
ALGEBRA 1 LESSON 11-5
Solve each radical equation.
x – 3 = x – 2 = x + 2
x + 7 = 5x – x = 2x + 8
x = 3 2 5 7 2 5 4
no solution
11-5