Alg1 RBC Answers A - Ms. Calvo's Sitelcalvo.weebly.com/uploads/5/8/6/2/58623473/unit_9_answers.pdfwhen the constant is zero, there is one solution; and when the constant is negative,
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The parabola crosses the x-axis twice. The points are
( )1, 0− and ( )3, 0 .
The zeros of the function show the same points because a “zero” is another name for a point where the parabola crosses the x-axis. These points are called zeros because they represent the point(s) on the parabola where the y-value is zero.
Replacing ( )f x with zero and moving the constant
yields the following equations:
2 4x = 2 9x = 2 0x = 2 0x = 2 4x = −
2 9x = −
When the constant is positive, there are two solutions; when the constant is zero, there is one solution; and when the constant is negative, there are no solutions.
9.3 Warm Up
1. 5x = 2. 6x =
3. 2w = − 4. 2a = −
9.3 Cumulative Review Warm Up
1. even 2. neither
3. neither 4. odd
9.3 Practice A
1. two; 6, 6x = − 2. zero
3. one; 0x = 4. 3, 3x = − 5. no solution
6. no solution 7. 8, 8x = − 8. 6, 6x = −
9. 0x = 10. 4x = 11. 5,1x = −
12. 8
, 23
x = − 13. 2.45x = ± 14. 4.24x = ±
15. 2.24x = ±
16. The square root does not distribute over subtraction,
so 2 9x − does not equal 3.x − Begin by
isolating 2x on the left side.
2
2
9 16
25
5
x
x
x
− =
== ±
17. a. 221 336w =
b. width 4 cm, length 12 cm
18. At 5,x = ± because 2 25.x =
19. At 1.1,x = ± because 2 1.21.x =
9.3 Practice B
1. two; 11,11x = − 2. zero
3. two; 14,14x = − 4. no solution
5. 2, 2x = − 6. 0x = 7. 3, 3x = −
8. 1 1,
3 3x = − 9.
9 9,
2 2x = − 10. 12, 2x = −
11. 1
, 22
x = − 12. 2 12
,5 5
x = −
13. 2.65x = ± 14. 2.83x = ± 15. 1.11x = ±
16. The right side becomes negative after subtracting 25, so the equation does not have a real solution.
The minimum and the vertex are the same point for the function.
A quadratic function has a minimum when the parabola opens up and a maximum when the parabola opens down; these correspond to positive and negative leading coefficients, respectively.
When 0,a = the Quadratic Formula is undefined because you cannot divide by zero. In this case, the function has no 2x term, and is therefore not quadratic but linear.
The other situation that makes the Quadratic Formula undefined is when 2 4 0.b ac− < Note: In this case, the equation is technically still solvable, using the set of imaginary numbers.
9.5 Warm Up
1. 23 2. 361− 3. 477
4. 191 5. 39− 6. 0.699
9.5 Cumulative Review Warm Up
1. 2 or 7w w≤ ≥
2. 5 1
3 3u− < < −
3. 4 0f− < <
4. 1 9
or5 5
v v≤ ≥
5. 7 17x− < <
6. no solution
9.5 Practice A
1. 2 5 0;x x+ = 1,a = 5,b = 0c =
2. 2 3 10 0;x x+ + = 1,a = 3,b = 10c =
3. 25 7 2 0;x x− − + = 5,a = − 7,b = − 2c =
4. 3x = − 5. no real solution
6. 1, 10x = − 7. 1
1,3
x = −
8. no real solution 9. 1
2x = −
10. 2 3
,3 2
x = − 11. 0.3, 2.1x = −
12. a. 0.3 and 0.8 sec b. 1.125 sec
13. one 14. two 15. zero
16. two 17. zero 18. two
19. 2;x = ± Using Square Roots; in the form 2x d=
20. no real solutions; Quadratic Formula; doesn’t fit the other methods
21. 0.4, 8.4;x = − completing the square; 1a = and
b is even.
22. 3.4, 2.4;x = − Quadratic Formula; doesn’t fit the
other methods
23. 7;x = factoring; Perfect Square Trinomial
24. 0, 5;x = factoring; easy to factor
25. a. 47 0;− < no real solutions
b. 23 5 6 0x x− + =
c. no; Both equations have zero solutions; the sign of b does not affect the discriminant because b is squared.
21. no real solution; using square roots; in the form 2x d=
22. 0.1,x = 1.4;− Quadratic Formula; doesn’t fit the
other methods
23. 1.1,x = − 1.9; Quadratic Formula; doesn’t fit the
other methods
24. 6;x = factoring; Perfect Square Trinomial
25. a. 47 0− < b. 23 5 6 0x x+ − =
c. yes; The original equation has zero solutions and the new equation has two solutions; changing the sign of c changes the discriminant from negative ( )47− to positive ( )97 .
9.5 Enrichment and Extension
1. 9 in.
2. length: 9 km, width: 12 km
3. width: 12 in., height: 16 in.
4. base: 22 4.7 in.,≈ height: 3 22 14.1 in.≈
5. height: 8 10 25.3 cm, base: 10 10 31.6 cm≈ ≈
9.5 Puzzle Time
THEIR TRUNKS
9.6 Start Thinking
System 1 has two solutions because the graphs of the equations intersect at two points.
System 2 has one solution because the graphs of the equations intersect at one point.
System 3 has no solution because the graphs of the equations do not intersect.
88. ( )2, 2− and ( )4, 10− 89. ( )2, 4− − and ( )1, 1−
90. no solution 91. 3 in. by 12 in.
Chapter 10 10.1 Start Thinking
The domain of the function is all positive real numbers and zero. The range is also all positive real numbers and zero. You cannot use negative numbers in the domain because the square root function cannot be evaluated using negative numbers in this course.
The calculator is able to graph this function because negative numbers and zero will allow the radical to be evaluated. This function has an inverse domain when compared to .y x=