Simplifying Radical Expressions

8.2. SIMPLIFYING RADICAL EXPRESSIONS
8.2
471
Simplifying Radical Expressions
√
√ √
1. Use the property a b = ab to
multiply the radicals.
√ √
5 13 = (5)(13)
√
= 65
√
√ √
3. Use the property a b = ab to
multiply the radicals.
√ √
17 2 = (17)(2)
√
= 34
√
√ √
5. Use the property a b = ab to
multiply the radicals.
√ √
5 17 = (5)(17)
√
= 85
7. From
9. From
Check: Use the graphing calculator to check the result.
Check: Use the graphing calculator to check the result.
Check: Use the graphing calculator to check the result.
√
√
56, we can factor out a perfect square, in this case 4.
√
√ √
56 = 4 14
Factor out a perfect square.
√
√
Simplify: 4 = 2.
= 2 14
√
√
99, we can factor out a perfect square, in this case 9.
√
√ √
99 = 9 11
Factor out a perfect square.
√
√
Simplify: 9 = 3.
= 3 11
Second Edition: 2012-2013
CHAPTER 8. QUADRATIC FUNCTIONS
472
11. From
13. From
15. From
17. From
19. From
21. From
23. From
√
√
150, we can factor out a perfect square, in this case 25.
√
√ √
150 = 25 6
Factor out a perfect square.
√
√
Simplify: 25 = 5.
=5 6
√
√
40, we can factor out a perfect square, in this case 4.
√ √
√
40 = 4 10
Factor out a perfect square.
√
√
= 2 10
Simplify: 4 = 2.
√
√
28, we can factor out a perfect square, in this case 4.
√ √
√
28 = 4 7
Factor out a perfect square.
√
√
Simplify: 4 = 2.
=2 7
√
√
153, we can factor out a perfect square, in this case 9.
√
√ √
153 = 9 17
Factor out a perfect square.
√
√
= 3 17
Simplify: 9 = 3.
√
√
50, we can factor out a perfect square, in this case 25.
√ √
√
50 = 25 2
Factor out a perfect square.
√
√
=5 2
Simplify: 25 = 5.
√
√
18, we can factor out a perfect square, in this case 9.
√
√ √
18 = 9 2
Factor out a perfect square.
√
√
Simplify: 9 = 3.
=3 2
√
√
44, we can factor out a perfect square, in this case 4.
√
√ √
44 = 4 11
Factor out a perfect square.
√
√
= 2 11
Simplify: 4 = 2.
Second Edition: 2012-2013
8.2. SIMPLIFYING RADICAL EXPRESSIONS
25. From
473
√
√
104, we can factor out a perfect square, in this case 4.
√ √
√
104 = 4 26
Factor out a perfect square.
√
√
= 2 26
Simplify: 4 = 2.
27. First, write out the Pythagorean Theorem, then substitute the given values
in the appropriate places.
a2 + b 2 = c2
2
2
Pythagorean Theorem.
2
(14) + b = (16)
2
Substitute: 14 for a, 16 for c.
Square: (14)2 = 196, (16)2 = 256.
196 + b = 256
b2 = 60
Subtract 196 from both sides.
√
√
The equation b2 = 60 has two real solutions, b = − 60 and b = 60. However,
in this situation, b represents the length of one leg of the right triangle and must
be a positive number. Hence:
b=
√
60
Nonnegative square root.
However, this answer is√not in simple radical form. In this case, we can factor
out the perfect square 4.
√ √
4 15
√
b = 2 15
b=
√
√ √
60 = 4 15.
√
4 = 2.
√
Thus, the length of the missing leg is b = 2 15.
29. First, write out the Pythagorean Theorem, then substitute the given values
in the appropriate places.
a2 + b 2 = c2
2
2
Pythagorean Theorem.
2
(3) + b = (25)
2
9 + b = 625
b2 = 616
Substitute: 3 for a, 25 for c.
Square: (3)2 = 9, (25)2 = 625.
Subtract 9 from both sides.
√
√
The equation b2 = 616 has two real solutions, b = − 616 and b = 616.
However, in this situation, b represents the length of one leg of the right triangle
and must be a positive number. Hence:
b=
√
616
Nonnegative square root.
Second Edition: 2012-2013
CHAPTER 8. QUADRATIC FUNCTIONS
474
However, this answer is√not in simple radical form. In this case, we can factor
out the perfect square 4.
√
√ √
√ √
616 = 4 154.
b = 4 154
√
√
4 = 2.
b = 2 154
√
Thus, the length of the missing leg is b = 2 154.
31. First, write out the Pythagorean Theorem, then substitute the given values
in the appropriate places.
a2 + b 2 = c2
2
2
(2) + (12) = c
4 + 144 = c
2
2
148 = c2
Pythagorean Theorem.
Substitute: 2 for a, 12 for b.
Square: (2)2 = 4, (12)2 = 144.
Add: 4 + 144 = 148.
√
√
The equation c2 = 148 has two real solutions, c = − 148 and c = 148.
However, in this situation, c represents the length of the hypotenuse and must
be a positive number. Hence:
√
Nonnegative square root.
c = 148
However, this answer is√not in simple radical form. In this case, we can factor
out the perfect square 4.
√ √
√
√ √
c = 4 37
148 = 4 37.
√
√
c = 2 37
4 = 2.
√
Thus, the length of the hypotenuse is c = 2 37.
33. First, write out the Pythagorean Theorem, then substitute the given values
in the appropriate places.
a2 + b 2 = c2
2
2
(10) + (14) = c
100 + 196 = c
2
2
296 = c2
Pythagorean Theorem.
Substitute: 10 for a, 14 for b.
Square: (10)2 = 100, (14)2 = 196.
Add: 100 + 196 = 296.
√
√
The equation c2 = 296 has two real solutions, c = − 296 and c = 296.
However, in this situation, c represents the length of the hypotenuse and must
be a positive number. Hence:
√
Nonnegative square root.
c = 296
Second Edition: 2012-2013
8.2. SIMPLIFYING RADICAL EXPRESSIONS
475
However, this answer is√not in simple radical form. In this case, we can factor
out the perfect square 4.
√
√ √
√ √
296 = 4 74.
c = 4 74
√
√
4 = 2.
c = 2 74
√
Thus, the length of the hypotenuse is c = 2 74.
35. To find the area of the shaded region, subtract the area of the triangle
from the area of the semicircle.
Area of shaded region = Area of semicircle − Area of triangle
To find the area of the semicircle, we’ll need to find the radius of the circle.
Note that the hypotenuse of the right triangle ABC is the diameter of the
circle. We can use the Pythagorean Theorem to find its length.
(AB)2 = (AC)2 + (BC)2
(AB)2 = (4)2 + (3)2
(AB)2 = 16 + 9
(AB)2 = 25
Hence, the diameter of the semicircle is AB = 5. The radius of the semicircle
is 1/2 of the diameter. Hence, the radius of the semicircle is r = 5/2. The area
of a full circle is given by the formula A = πr2 , so the area of the semicircle is
found by:
1 2
πr
2
2
5
1
= π
2
2
25
π
=
8
To find the area of ABC, we take half of the base times the height.
Area of semicircle =
1
(3)(4)
2
=6
Area of triangle =
We can now find the area of the shaded region.
Area of shaded region = Area of semicircle − Area of triangle
25
=
π−6
8
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CHAPTER 8. QUADRATIC FUNCTIONS
476
37. We follow the Requirements for Word Problem Solutions.
1. Set up a variable dictionary. Let x represent the length of the shorter
leg of the right triangle. Because the longer leg is 10 feet longer than
twice the length of the shorter leg, its length is 2x + 10. The length of
the hypotenuse is 4 feet longer than three times the length of its shorter
leg, so its length is 3x + 4. A sketch will help us maintain focus.
3x + 4
x
2x + 10
2. Set up an equation. By the Pythagorean Theorem, the sum of the squares
of the legs must equal the square of the hypotenuse.
x2 + (2x + 10)2 = (3x + 4)2
3. Solve the equation. Use the shortcut (a + b)2 = a2 + 2ab + b2 to expand.
x2 + (2x + 10)2 = (3x + 4)2
x2 + 4x2 + 40x + 100 = 9x2 + 24x + 16
Simplify the left-hand side.
5x2 + 40x + 100 = 9x2 + 24x + 16
The equation is nonlinear, so make one side zero by subtracting 5x2 , 40x,
and 100 from both sides of the equation.
0 = 4x2 − 16x − 84
Note that each coefficient is divisible by 4. Divide both sides of the
equation by 4.
0 = x2 − 4x − 21
Note that the integer pair 3 and −7 has a product equal to ac = (1)(−21) =
−21 and sum equal to b = −4. Because the leading coefficient is a 1, we
can “drop this pair in place” to factor.
0 = (x − 7)(x + 3)
Hence, x = 7 and x = −3.
Second Edition: 2012-2013
8.2. SIMPLIFYING RADICAL EXPRESSIONS
477
4. Answer the question. Because x represents the length of the shortest side
of the right triangle, and length is a positive quantity, we discard the
answer x = −3. Next, if the length of the shorter leg is x = 7 feet, then
the length of the longer leg is 2x + 10 = 2(7) + 10, or 24 feet. Finally, the
length of the hypotenuse is 3x + 4 = 3(7) + 4, or 25 feet.
5. Look back. Note that the length of the longer leg is 24 feet, which is 10
feet longer than twice 7 feet, the length of the shorter leg. The length of
the hypotenuse is 25, which is 4 feet longer than three times 7 feet, the
length of the shorter leg. But do the numbers satisfy the Pythagorean
Theorem?
25
7
24
That is, is it true that
?
72 + 242 = 252
Simplify each square.
?
49 + 576 = 625
Note that this last statement is true, so our solution checks.
39. As always, we obey the Requirements for Word Problem Solutions.
1. Set up a variable dictionary. We’ll create a well-marked diagram for this
purpose, letting h represent the distance between the base of the garage
wall and the upper tip of the ladder.
Second Edition: 2012-2013
CHAPTER 8. QUADRATIC FUNCTIONS
478
19 ft
h
5 ft
2. Set up an equation. Using the Pythagorean Theorem, we can write:
52 + h2 = 192
Pythagorean Theorem.
2
Square: 52 = 25 and 192 = 361.
25 + h = 361
3. Solve the equation.
h2 = 336
√
h = 336
Subtract 25 from both sides.
h will be the nonnegative square root.
4. Answer the question. As we’re asked for a decimal approximation, we
won’t worry
about simple radical form in this situation. The ladder
√
reaches 336 feet up the wall. Using a calculator, this is about 18.3 feet,
rounded to the nearest tenth of a foot.
5. Look back. Understand that when we use 18.3 ft, an approximation, our
solution will only check approximately.
19 ft
5 ft
Second Edition: 2012-2013
18.3 ft