Week 1 Assignment 4: Mixed Problems
Math 101 Quiz Week 5
Submit your answers directly to the facilitator by Day 6; please show all necessary work.
For questions 1-2, apply the quadratic formula to find the roots of the given function, and then graph the function.
1. (6 points) f(x) = x2 + 4
X = (0 +/- sqrt(0^2-4*1*4))/2
X = (+/- sqrt(-16))/2
X = +/- i sqrt(16))/2
X = +/- 2i
X = 2i or -2i
[pic]
2. (6 points) g(x) = x2 + x + 12
X = (-1 +/- sqrt(1^2-4*1*12))/2
X = (-1 +/- sqrt(-47))/2
X = -1/2 – sqrt(47)/2 i or -1/2 + sqrt(47)/2 i
[pic]
For questions 3-4, factor the quadratic expression completely, and find the roots of the expression.
3. (4 points) 135x2 - 222x + 91
(15x-13)(9x-7) = 0
X = 13/15 or 7/9
4. (4 points) 6x2 - 42x + 72
6(x^2-7x+12) = 0
6(x-4)(x-3) = 0
x = 3 or 4
For question 5, complete the square, and find the roots of the quadratic equation.
5. (3 points) x2 - 2/3x = 0
x^2 – 2/3x + 1/9 = 1/9
(x-1/3)^2 = 1/9
x-1/3 = +/- 1/3
x = 1/3 +/- 1/3
x = 0 or 2/3
In questions 6-10, use the discriminant to determine the number of solutions of the quadratic equation, and whether the solutions are real or complex. Note: It is not necessary to find the roots; just determine the number and types of solutions.
6. (3 points) 2x2 + x - 1 = 0
1^2-4*2*-1 = 9, two real
7. (3 points) 4/3x2 - 2x + 3/4 = 0
2^2-4*4/3*3/4 = 0, one real
8. (3 points) 2x2 + 5x + 5 = 0
5^2-4*2*5 = -15, two complex
9. (3 points) 3z2 + z - 1 = 0
1^2-4*3*-1 = 13, two real
10. (3 points) m2 + m + 1 = 0
1^2-4*1*1 = -3, two complex
11. (5 points) A rectangular parking lot is 50 ft longer than it is wide. Determine the dimensions of the parking lot if it measures 250 ft diagonally.
Pythagorean Theorem:
x^2 + (x+50)^2 = 250^2
x^2 + x^2 + 100x + 2500 = 62500
2x^2 + 100x – 60000 = 0
x^2 + 50x – 30000 = 0
(x+200)(x-150) = 0
x = -200 or 150
Can’t be negative, so:
X = 150 feet
X+50 = 200 feet
They are 150 ft and 200 ft.
12. (5 points) Three consecutive even integers are such that the square of the third is 76 more than the square of the second. Find the three integers.
(x+4)^2 = (x+2)^2 + 76
x^2 + 8x + 16 = x^2 + 4x + 4 + 76
8x + 16 = 4x + 80
4x = 64
x = 16
They are 16, 18, 20
13. (5 points) A photo is 3 inches longer that it is wide. A 2-inch border is placed around the photo making the total area of the photo and border 108in2. What are the dimensions of the photo?
The original photo is x by (x+3), but we add 2 to each dimension and set it to 108:
(x+2)(x+2+3) = 108
(x+2)(x+5) = 108
FOIL:
x^2 + 2x + 5x + 10 = 108
x^2 + 7x + 10 = 108
x^2 + 7x - 98 = 0
Factor:
(x-7)(x+14) = 0
x = 7 or -14
You can't have a negative side, so x = 7. If one side is 7, the other side is 3 longer.
The original was 7 inches by 10 inches.
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