Download CHAPTER 7 – Selected Answers “Evens”

Algebra 2 - CHAPTER 7 – Selected Answers “Evens”
p.1
Section 7.1 (p.309)
Oral Ex/ 1. 5, 3 2. 2, –8 3. 2, –1 4.  74 ,  94 5. 11  2, 11  2 6. 6  6,  6  6
7. 6  3i, 6  3i 8.
4
3

2
i, 34
3

2
i
3
9a.  5 9b. x  3   5 9c. x  3   5
9d. 2x  3   5 10a. y   1 b. y  2   1 c. y  2   1 d. 6 y  2   1
1 19. 1 20. 0, 6 – factoring
11. 1 12. 4 13. 16 14. 36 15. ¼ 16. 94 17. 49
18. 25
4
16
Written Exercises
4  6
3 5i
4a.  7 b. 15, 1 c. 3, 1/5 6a.  5 i b. 3  5 i c.
3
5
8. 3  2 2 10. 24  10 2 12. 5  2 6 i 14. 2  2 16. 4  10 18. 5  5 i
2a.  6 b. 4  6 c.
20. 2  2i 22. 2 
33
3
24. 2 
34. 1  9 2 36. 1 38. 0 40.
22
2
26.
3  29
1  71
28. 3  13 30. 3  9i 32.
7
2
1  1  4c
2
Section 7.2 (p.313)
Oral Ex/ 1. a = 2, b = –3, c = 7 2. a = 3, b = 7, c = –2 3. a = –4, b = –7, c = 5
4. a = 1, b = 2, c = –4 5. a = 1, b = – 5 , c = 1 6. a = 1, b = –2, c = –9
Written Exercises
3  29
3  19
1 2
2.
4. 3  3 6. 2, –1/2 8. 1  22 i 10. 4  17 12.
14.
2
4
2
7  52
16. 15  75 i 18.
20. –1.27, 2.77 22. –1.47, 0.47 24. 0.42, 3.58 26. 4
2
5 3
28. -5/2, 3/2 30.
32.  3  6 34. 33  315 i 36. –i 38. i, 3 + i 40.  13 i, i
2
b  b2  4ac
42. By the quadratic formula, the roots of ax + bx + c = 0 are
.
2a
2
b  b2  4ac
b  b2  4ac
Let x1 =
and x2 =
. Suppose x1 and x2 are reciprocals.
2a
2a
That is, if x2 = x1 , then x1x2 = 1 (ie, their product is equal to ‘1’). Multiply these two roots
1
as a ‘difference of two squares’ to get: x1x2 =
4ac
c

1

 1  c  a q.e.d .
a
4a 2

b 2  b 2  4ac
4a 2
 and set this equal to ‘1’ as:
CHAPTER 7 – Selected Answers “Evens”
Section 7.2 (continued) – Problems
2. 4 4. 1.38, 3.62 6. 20 m by 60 m, 30 m by 40 m 8. 1.21 in 10. 5.31
1 5
12.
14. 14.47 cm on a side
2
p.2
Section 7.3 (p.319)
Oral Ex/ 1. 24(irrational roots) 2. 16(rational roots) 3. 0(one root, rational)
4. –4(imaginary roots) 5. 45(irrational roots) 6. –8(imaginary roots)
7. 25(irrational roots) 8. 16(irrational roots) 9. 16(irrational roots)
10. By ‘equal’ roots, the text means that there is just one root (repeated). This would mean
that the discriminant b2 – 4ac = 0, but we’re given that b2 – 4ac > 0 (positive, not zero).
Written Exercises
2. rational roots 4. irrational roots 6. imaginary (conjugate) roots 8. one rational root
(often called a ‘repeated’ or ‘double’ root) 10. rational roots 12. one irrational root
14. –6, 4 16. –3/2 18. 7  5 20. 3  i 22. –22, 18 24. –3, 7/2 26. 6  2 6 28.  22
30. 4/3, 3 32a. k = –3 b. k > –3 c. k < –3 34a. k = 1 b. –1 < k < 1 c. k < –1 or k > 1
36a. k = 2 b. k < –2 or k > 2 c. –2 < k < 2 38. Sample Answer: 0, 3, –9
40a. 13  13 i b. Adding we get 2/3, multiplying (as a difference of 2 squares) we get: 19  19
b
7
c 8
c. For 3x2 + 7x + 8 = 0, a = 3, b = 7, c = 8, Sum:  =  , Product:  (see below)
a
3
a 3
Sum and Product of the Roots of a Quadratic
Theorem: For ax2 + bx + c = 0, the sum of the roots is: b/a the product of the roots is c/a.
Proof:
Let x1 =
2b
b
b  b2  4ac
b  b2  4ac
and x2 =
. Then x1 + x2 =
  and
2a
a
2a
2a

b 2  b 2  4ac
the product is (as a difference of 2 squares): x1x2 =
 = 4ac  c
q.e.d .
4a 2
4a 2 a
42. For x2 + kx + (k – 1) = 0, the discriminant is: k2 – 4(1)(k – 1) = k2 – 4k + 4 and this
equals (k – 2)2 which is always a perfect square for k that’s an integer. Hence rational roots.
Section 7.4 (p.323)
Oral Ex/ 1. Let u = x + 2, then u2 – 5u – 14 = 0 2. Let u = 3x + 4, then u2 + 6u – 16 = 0
1 , then u2 – 4u – 5 = 0 4. Let u = 2/x, then u2 + 5u – 24 = 0
3. Let u = 3x
5. Let u = x2, then u2 – 10u + 9 = 0 6. Let u = x3, then u2 – 9u + 8 = 0
7. Let u = x , then 2u2 – 3u + 1 = 0 8. Let u = |x|, then 7u2 – 20u – 3 = 0
9. Let u = x2 – 5, then u2 – u – 2 = 0 10. Let u = x x 2 , then 5u2 – 3u – 2 = 0
11. Let u = x-1, then 2u2 + 3u – 2 = 0 12. Let u = x-1, then 3u2 – 4u + 1 = 0
CHAPTER 7 – Selected Answers “Evens”
p.3
Section 7.4 (continued)
Written Exercises
2a. 11, 16 b. –8/3, –1 c. 2,  3 4a. 4, 12 b. 2, 
2 6
c. 12 ,
3
3
2
6a. 1,  2 2 i b. 1
2
2
,  2 i b. 1 
, 1  2 i 10. 9 12. 2,  2i 14. 0, 3/2 16. 1/9 18. 1, 9 20. 1, 9/4
2
2
22. 3.30 24. 0.14 26. –5.24,–5,–1,–0.76 28. 4(x – 1)4 – 55(x – 1)2 + 39=0 30. 3 32. 4, 9
8a. 
Section 7.5 (p.330)
Oral Ex/ 1. c 2. e 3. h 4. b 5. a 6. g 7. f 8. d 9a. down b. (0,0) c. x = 0
10a. up b. (0, –1) c. x = 0 11a. down b. (1,0) c. x = 1 12a. up b. (5,4) c. x =5
13a. down b. (2, –3) c. x = 2 14a. up b. (–1, –2) c. x = –1 15. right 1 16. down 4
17. right 1, down 4
Written Exercises
CHAPTER 7 – Selected Answers “Evens”
p.4
Sorry, but the equations will be written in the y = f(x) or completed square form! (See the
next section and page 333.)
20. y = -2(x – 4)2 + 5 22. y   23 ( x  3)2  6 24. y  2( x  2)2  6
26. y  1( x  3) 2  4 28. 3 30. 9/4 32. a = ½ , k = –1
38a. V(0,0), y = 0 b. V(2,0), y = 0 c. V(0,-1), y = -1 d. V(2,-1), y = -1 e. V(3,2), y = 2
Section 7.6 (p.335)
Oral Ex/ 1. min, x = 0 2. max, x = 0 3. min, x = 1 4. max, x = 1 5. min, x=5 6. min, x=5
7a. 0 b. 0
Written Exercises
CHAPTER 7 – Selected Answers “Evens”
p.5
20. min, y = -10, V(-4,-10) 22. min, y = -49/12, V(5/6, -49/12) 24. min y = -16 V(1/2,-16)
26. V(1,3), D = Reals, R: y  3 , x = 0, 2 28. V(-4,25), D = Reals, R: y  25 , x = -9, 1
4  6
32. V(-5,-1/2), D=Reals, R: y   12 , x = -6,-4
2
34. V(5,-8), D=Reals, R: y  8 , x = 3, 7 36. V(1,-1/2), D=Reals, R: y   12 , x = 23 , 43
38a. positive b. zero c. negative d. positive 40a. 0 b. 1 c. 2
2
4ac  b 2
42. y 
 a x  2ab 44. Show f(x) = ax2 + bx + c is never odd (if a and b are
4a
not zero!). f(-x) = a(-x)2 + b(-x) + c = ax2 – bx + c and –f(x) = –ax2 – bx – c. So to be odd,
these two expressions must be equal for all real x’s. If f(-x) = -f(x) then 2ax2 = –2c or
x2 = –c/a which can be true for at most 2 values of x (not for all x). Hence f(x) is not odd.
30. V(-2,-3), D=Reals, R: y  3 , x=


Section 7.7 (p.341)
Oral Ex/ 1. 3, 6 2. 6, 8 3. ½ , -3 4. 3, ½ 5. 3/2, 1 6. 0, -4/3 Sample Answers (follow):
7. x2 – 3x + 4 = 0 8. x2 + 2x – 2 = 0 9. x2 – 1 = 0 10. x2 + 4x = 0 11. x2 – x – 6 = 0
12. x2 – 10x + 9 = 0 13. x2 – 7x = 0 14. x2 – 5 = 0 15. The product of the roots should be
c/a or 3/2, but the for 2y2 – 7y + 3 = 0, if 2 and 3/2 are the roots, their product is 2(3/2) = 3.
Written Exercises
2 3
3 7
2. 3, 4 4.
6. 
8. x2 + 2x – 3 = 0 10. 4x2 – 4x – 3 = 0 12. 4x2 – 5 = 0
3
2
2
14. x – 4x – 3 = 0 16. 16x2 + 16x – 1 = 0 18. x2 + 8 = 0 20. x2 – 8x + 20 = 0
22. x2 + 4x + 11 = 0 24. 4x2 – 8x + 7 = 0 26. f ( x)   23 x 2  34 x  16
3
28. f ( x)  54 x 2  54 x  24
30. f ( x)  52 x 2  54 x  48
32. f(x) = -x2 – 6x
5
5
34. f ( x)  94 x 2  83 x 36. f(x) = -x2 + 2x + 8 38. Show? Well, plug it in and you should get:
23  10 2 i  50  10 2 i  27 = 0 40. See Section 7.3 answers (above) for the theorem
about the sum (r1 + r2 = -b/a) and product (r1r2 = c/a) of the roots of a quadratic equation.
1 1
b
Sum of the Reciprocals Theorem -   
r1 r2
c
Proof: Adding with LCD of r1r2 :
r2
r2

1 1 r1 r2  r1
  

r1 r2 r1
r1r2
b
c
a
a

b
q.e.d .
c
Problems (p.343)
2. x = 8, x2 + 8x, -16 4. 10, 10 6. 1250 m2 8. 28, $270.40 10. $15
12a. 100ft b. 5 sec after being thrown 14. 78 m by 78 m 16. 20
Chapter 7 Review 9 (p.346)
2. b 4. b 6. d 8. a 10. d 12. c 14. c