Download Math 380 Spring 2016 T. Matsumoto Sec 8.1 Practice Radical

Math 380
Spring 2016
T. Matsumoto
Sec 8.1 Practice Radical Notation
NAME:
Instructions: For multiple choice questions, write the letter corresponding to your answer in the
blank provided. For other questions, show all your work in the space provided, and write your
answer in the blank. All answers must be exact unless otherwise indicated.
1.
(1pt ) List all real square roots of 169.
[1]
2.
(1pt ) List all real square roots of 256.
[2]
3.
(1pt ) List all real square roots of 196.
[3]
4.
(1pt ) List all real solutions of x2 = 36.
[4]
5.
(1pt ) List all real solutions of x2 = 625.
[5]
6.
(1pt ) List all real solutions of x2 = 49.
[6]
7.
√
(1pt ) Simplify: −576
(a)
The expression is not a real number.
(b)
(c)
−24
24
[7]
8.
√
(1pt ) Simplify: − 49
[8]
9.
√
(1pt ) Simplify: − 441
� √ �2
10. (1pt ) Simplify: − 61
� √ �2
11. (1pt ) Simplify: − 47
� √ �2
12. (1pt ) Simplify: − 2
[9]
[10]
[11]
[12]
13. (1pt ) Use the 5:intersect utility on CALC menu of the graphing calculator to solve the equation x2 = 30. Round your answers to the
nearest tenth.
[13]
14. (1pt ) Use the 5:intersect utility on CALC menu of the graphing calculator to solve the equation x2 = 37. Round your answers to the
nearest tenth.
[14]
15. (1pt ) Use the 5:intersect utility on CALC menu of the graphing calculator to solve the equation x2 = 29. Round your answers to the
nearest tenth.
[15]
Answers
1. -13,13
2. -16,16
3. -14,14
4. -6,6
5. -25,25
6. -7,7
7. a
8. -7
9. -21
10. 61
11. 47
12. 2
13. -5.5,5.5
14. -6.1,6.1
15. -5.4,5.4
Solutions
1.
We are looking for a number whose square is 169.
• Because (−13)2 = 169, the negative square root of 169 is −13.
• Because (13)2 = 169, the nonnegative square root of 169 is 13.
Hence, 169 has two real square roots, −13 and 13.
2.
We are looking for a number whose square is 256.
• Because (−16)2 = 256, the negative square root of 256 is −16.
• Because (16)2 = 256, the nonnegative square root of 256 is 16.
Hence, 256 has two real square roots, −16 and 16.
3.
We are looking for a number whose square is 196.
• Because (−14)2 = 196, the negative square root of 196 is −14.
• Because (14)2 = 196, the nonnegative square root of 196 is 14.
Hence, 196 has two real square roots, −14 and 14.
4.
There are two numbers whose square equals 36, namely −6 and 6. Hence:
x2 = 36
x = ±6
Original equation.
Two answers: (−6)2 = 36 and (6)2 = 36.
Thus, the real solutions of x2 = 36 are x = −6 or x = 6. Writing x = ±6 (“x equals plus or
minus 6) is a shortcut for writing x = −6 or x = 6.
5.
There are two numbers whose square equals 625, namely −25 and 25. Hence:
x2 = 625
x = ±25
Original equation.
Two answers: (−25)2 = 625 and (25)2 = 625.
Thus, the real solutions of x2 = 625 are x = −25 or x = 25. Writing x = ±25 (“x equals plus
or minus 25) is a shortcut for writing x = −25 or x = 25.
6.
There are two numbers whose square equals 49, namely −7 and 7. Hence:
x2 = 49
x = ±7
Original equation.
Two answers: (−7)2 = 49 and (7)2 = 49.
Thus, the real solutions of x2 = 49 are x = −7 or x = 7. Writing x = ±7 (“x equals plus or
minus 7) is a shortcut for writing x = −7 or x = 7.
7.
8.
9.
√
The expression −576 calls for the nonnegative square
root of −576. Because you cannot
√
square a real number and get −576, the expression −576 is not a real number.
√
The expression − 49 calls for the negative square root of 49. Because (−7)2 = 49 and −7 is
negative,
√
− 49 = −7.
√
The expression − 441 calls for the negative square root of 441. Because (−21)2 = 441 and
−21 is negative,
√
− 441 = −21.
√
√
10. If a > 0, then − a is the negative solution of x2 = a. Hence, when we substitute − a into
the equation x2 = a, we must get a true statement:
√
(− a)2 = a.
Thus:
√
(− 61)2 = 61.
√
√
11. If a > 0, then − a is the negative solution of x2 = a. Hence, when we substitute − a into
the equation x2 = a, we must get a true statement:
√
(− a)2 = a.
Thus:
√
(− 47)2 = 47.
√
√
12. If a > 0, then − a is the negative solution of x2 = a. Hence, when we substitute − a into
the equation x2 = a, we must get a true statement:
√
(− a)2 = a.
Thus:
√
(− 2)2 = 2.
13. Enter each side of the equation x2 = 30 in the Y= menu, then adjust the WINDOW settings so
that the points of intersection are visible in the viewing window. Use the 5:intersect utility on
the CALC menu to determine the coordinates of the point of intersection. Report your results
as shown in the following figure.
y
40
y = x2
y = 30
−10 −5.5
5.5
10
x
−40
Hence, to the nearest tenth, the solutions of x2 = 30 are x = −5.5 or x = 5.5.
14. Enter each side of the equation x2 = 37 in the Y= menu, then adjust the WINDOW settings so
that the points of intersection are visible in the viewing window. Use the 5:intersect utility on
the CALC menu to determine the coordinates of the point of intersection. Report your results
as shown in the following figure.
y
50
y = x2
y = 37
−10 −6.1
6.1
10
x
−50
Hence, to the nearest tenth, the solutions of x2 = 37 are x = −6.1 or x = 6.1.
15. Enter each side of the equation x2 = 29 in the Y= menu, then adjust the WINDOW settings so
that the points of intersection are visible in the viewing window. Use the 5:intersect utility on
the CALC menu to determine the coordinates of the point of intersection. Report your results
as shown in the following figure.
y
40
y = x2
y = 29
−10 −5.4
5.4
10
x
−40
Hence, to the nearest tenth, the solutions of x2 = 29 are x = −5.4 or x = 5.4.