Tutors Answer Your Questions about Quadratic Equations (FREE)
Question 193342: We're doing Translation.
It asks What happens to the vertex in the following equations as they are translated?
A) y= |x| changes to y=|x+2|
I entered it into my graphing calculatorl, but how do I set it up into the question?
Click here to see answer by RAY100(1637) |
Question 193345: I have a problem that states to ADD THE PROPER CONSTANT TO EACH BINOMIAL SO THAT THE RESULTING TRINOMIAL IS A PERFECT SQUARE TRINOMIAL. THEN FACTOR EACH TRINOMIAL.
x^2-18x+____
How would you solve this?
any help would be greatly appreciated!!
Click here to see answer by Mathtut(3670) |
Question 193345: I have a problem that states to ADD THE PROPER CONSTANT TO EACH BINOMIAL SO THAT THE RESULTING TRINOMIAL IS A PERFECT SQUARE TRINOMIAL. THEN FACTOR EACH TRINOMIAL.
x^2-18x+____
How would you solve this?
any help would be greatly appreciated!!
Click here to see answer by RAY100(1637) |
Question 193340: Please check my work, Thanks.
Assignment 2: Quadratic Equations
Solve the following question on quadratic equations
Write three quadratic equations, with a, b, and c (coefficients of x2, x, and the constant) as:
1. Integers
2x^2 -3x + 4 =0
2. Rational numbers other than integers. (Integers will not count for this one.)
(1/2)x^2 +(3/5)x + (1/7) = 0
3. Irrational numbers
√2x^2 + √3x -√5 = 0
Click here to see answer by RAY100(1637) |
Question 193390: Please check my work, thank you.
Exponential Functions
Question 1:
Do exponential functions only model phenomena that grow, or can they also model phenomena that decay? Explain what is different in the form of the function in each case.
Yes, they can also model decay, for example you can use them to model the decay of radioactivity. The only difference is that there will be a negative index.
Example of function for growth: y=ae^(kt)
Example of function for decay: y=ae^(-kt)
Question 2:
A Cell divides into two identical copies every 4 minutes. How many cells will exist after 5 hours?
Growth formula:
A(t) = A(o)(2)^(t/4) where t is number of minutes after t = 0
------------------------------------------------------------------
A(300) = 1*2^(300/4)
A(300) = 2^(75)
A(300) = 3.77x10^22 cells
Click here to see answer by yuendatlo(23) |
Question 193387: Please check my work, thank you.
Practical Application of Quadratic Equations
1. A rectangular garden has dimensions of 15 feet by 11 feet. A gravel path of uniform width is to be built around the garden. How wide can the path be if there is enough gravel for 192 square feet?
Ag = area of garden = 15 * 11 = 165 sq feet
Ap = area of gravel path = (15 + 2x)(11 + 2x) - 165
Ap = 165 + 52x + 4x^2 - 165
Ap = 4x^2 + 5x
where
x = width of the gravel path around the garden
and since the available gravel is 192 sq ft, then the above equation becomes
4x2 + 5x = 192
and rewriting,
4x2 + 5x - 192 = 0
Using the quadratic formula,
x = 6.33 feet
ANSWER: The width of the gravel path should be about 6' 4".
2. A business invests $8,000 in a savings account for two years. At the beginning of the second year, an additional $2,500 is invested. At the end of the second year, the account balance is $11,445. What was the annual interest rate?
the annual interest rate = r
$8,000 in a savings account for two years.--> 8000*(1+r)^2
beginning of the second year, an additional $2,500 is invested-->2500*(100+r)
8000*(1+r)^2 + 2500*(1+r) = 11,445
suppose 100+r = A
8000*A^2 + 2500*A = 11,445
8000*A^2 + 2500*A - 11,445 = 0
A = 1.05
r = 0.05 or 5 % Ans.
3. Steve traveled 150 miles at a certain speed. Had he gone 20mph faster, the trip would have taken 2 hours less. Find the speed of his vehicle.
Let R = rate of Steve
150/R 150/(R + 20) = 2
150(R+ 20) 150(R) = 2(R)(R + 20)
150R + 3000 150R = 2Rē + 40R
0 = 2Rē + 40R 3000
Rē + 20R 1500 = 0
(R + 50)(R 30) = 0
(R + 50) = 0
R = 50 Discard, negative rate
(R 30) = 0
R = 30 mph (speed of vehicle)
4. The Hudson River flows at a rate of 5 miles per hour. A patrol boat travels 40 miles upriver, and returns in a total time of 6 hours. What is the speed of the boat in still water?
Let R = rate of patrol boat
40/(R 5) + 40(R + 5) = 6
40(R + 5) + 40(R + 5) = 6(R + 5)(R 5)
40R 200 + 40R + 200 = 6Rē 150
0 = 6Rē 80R 150
3Rē 40R 75 = 0
(3R + 5)(R 15) = 0
(3R + 5) = 0
3R = 5
R = 5/3 Discard, negative rate
(R 15) = 0
R = 15 mph (speed of boat in still water)
Click here to see answer by yuendatlo(23) |
Question 193381: I couldn't work letter C and #2, please assist.
Please check the rest of the work and let me know if it is correct.
Solutions of Quadratic Equations and their Applications
1. Determine whether the following equations have a real or complex solution. Justify your answer.
a) x2 + 3x - 15 = 0
a = 1
b = 3
c = -15
Discriminant: b2-4ac = 3 2-4*1*(-15) = 69
Discriminant (69) is greater than zero. The equation has two solutions
b) x2 + x + 4 = 0
a = 1
b = 1
c = 4
Discriminant: b2-4ac = 1 2-4*1*4 = -15
Discriminant (-15) is less than zero. No solutions are defined.
c)
d) x2 8x + 16 = 0
a = 1
b = -8
c = 16
Discriminant: b2-4ac = 8 2-4*1*16 = 0
Discriminant (0) is zero. There is only one solution.
e) 2x2 - 3x + 7 = 0
b = -3
c = 7
Discriminant: b2-4ac = 3 2-4*2*7 = -47
Discriminant (-47) is less than zero. No solutions are defined.
f) x2 4x - 77 = 0
a = 1
b = -4
c = -77
Discriminant: b2-4ac = 4 2+4*1*(-77) = 324
Discriminant (324) is greater than zero. The equation has two solutions
g) 3x2 - 7x + 6 = 0
a = 3
b = -7
c = 6
Discriminant: b2-4ac = 7 2-4*3*6 = -23
Discriminant (-23) is less than zero. No solutions are defined.
h) 4x2 + 16x + 16 = 0
a = 4x
b = 16
c = 16
Discriminant: b2-4ac = 16 2-4*4*16 = 0
Discriminant (0) is zero. There is only one solution.
2. Find an equation for which -3 and 4 are solutions.
3. What type of solution do you get for quadratic equations where D < 0? Give reasons for your answer. Also provide an example of such a quadratic equation and find the solution of the equation.
You get complex solutions for b^2-4ac<0.
x^2 +4x +5 = 0
x = [-4 +/-(16-20)^1/2]/2 = -2 +/-i
4. Create a real-life situation that fits the equation (x + 3)(x - 5) = 0 and express the situation as the same equation.
The length of a rectangular field is 2 feet more than the width and the area of the field is 15 sq feet.
Here is how we arrived at the equation from real life situation -
Let the length be x, then width = x-2,
and area = length * width
so. x(x-2) = 15
so, x^2 - 2x -15 =0
Factoring we get
(x+3)(x - 5) = 0
so, x=5 (as lenth has to be positive)
hence width = 3 feet
Click here to see answer by yuendatlo(23) |
Question 193381: I couldn't work letter C and #2, please assist.
Please check the rest of the work and let me know if it is correct.
Solutions of Quadratic Equations and their Applications
1. Determine whether the following equations have a real or complex solution. Justify your answer.
a) x2 + 3x - 15 = 0
a = 1
b = 3
c = -15
Discriminant: b2-4ac = 3 2-4*1*(-15) = 69
Discriminant (69) is greater than zero. The equation has two solutions
b) x2 + x + 4 = 0
a = 1
b = 1
c = 4
Discriminant: b2-4ac = 1 2-4*1*4 = -15
Discriminant (-15) is less than zero. No solutions are defined.
c)
d) x2 8x + 16 = 0
a = 1
b = -8
c = 16
Discriminant: b2-4ac = 8 2-4*1*16 = 0
Discriminant (0) is zero. There is only one solution.
e) 2x2 - 3x + 7 = 0
b = -3
c = 7
Discriminant: b2-4ac = 3 2-4*2*7 = -47
Discriminant (-47) is less than zero. No solutions are defined.
f) x2 4x - 77 = 0
a = 1
b = -4
c = -77
Discriminant: b2-4ac = 4 2+4*1*(-77) = 324
Discriminant (324) is greater than zero. The equation has two solutions
g) 3x2 - 7x + 6 = 0
a = 3
b = -7
c = 6
Discriminant: b2-4ac = 7 2-4*3*6 = -23
Discriminant (-23) is less than zero. No solutions are defined.
h) 4x2 + 16x + 16 = 0
a = 4x
b = 16
c = 16
Discriminant: b2-4ac = 16 2-4*4*16 = 0
Discriminant (0) is zero. There is only one solution.
2. Find an equation for which -3 and 4 are solutions.
3. What type of solution do you get for quadratic equations where D < 0? Give reasons for your answer. Also provide an example of such a quadratic equation and find the solution of the equation.
You get complex solutions for b^2-4ac<0.
x^2 +4x +5 = 0
x = [-4 +/-(16-20)^1/2]/2 = -2 +/-i
4. Create a real-life situation that fits the equation (x + 3)(x - 5) = 0 and express the situation as the same equation.
The length of a rectangular field is 2 feet more than the width and the area of the field is 15 sq feet.
Here is how we arrived at the equation from real life situation -
Let the length be x, then width = x-2,
and area = length * width
so. x(x-2) = 15
so, x^2 - 2x -15 =0
Factoring we get
(x+3)(x - 5) = 0
so, x=5 (as lenth has to be positive)
hence width = 3 feet
Click here to see answer by solver91311(16897)  |
Question 193440: i need some help on a word problem for quadratics-it goes like this-a model rocket is fired into the air off the top of a building. the following equation models this motion: h=-10t+26t+12 where h is the height in meters above the ground and t is the time in seconds since the launch. and now i have to answer the following questions...A. how high is the building? B. how high will the rocket be after 2.5 seconds? C. when will the rocket reach a height of 12 meters? D. how long will it take for the rocket to hit the ground? E. how long will it take to reach its maximum height? F. what is the maximum height the rocket can go?
i was able to figure out B and C but i am having a little bit of trouble figuring out the rest. i do not want answers i just need a little help getting started on the others. thank you sooo much!!!
Click here to see answer by Alan3354(30993)  |
Question 193573: Please check my work:
Form each of the following:
A linear equation in one variable: X + 3 = -5
A linear equation in two variables: y = 5x - 3
A quadratic equation: 3x2 + x - 2 = 0
A polynomial of three terms: 7x2y3 + 4x − 9
An exponential function: 32 = 9
A logarithmic function: log39 = 2
Click here to see answer by stanbon(57387) |
Question 193575: I'm sorry. I know this problem is very similar to the example, but I get a very crazy answer when I've tried to do this like the example many times. I must be doing something wrong. Please help.
One leg of a right triangle is 96 inches. Find the hypotenuse and teh other leg if the length of the hypotenuse exceeds 2 and 1/2 times teh other leg by 4 inches.
Click here to see answer by nerdybill(6963)  |
Question 193575: I'm sorry. I know this problem is very similar to the example, but I get a very crazy answer when I've tried to do this like the example many times. I must be doing something wrong. Please help.
One leg of a right triangle is 96 inches. Find the hypotenuse and teh other leg if the length of the hypotenuse exceeds 2 and 1/2 times teh other leg by 4 inches.
Click here to see answer by stanbon(57387) |
Question 193709: Translate the following into a quadratic equation, and solve it: The length of a rectangular garden is four times its width; if the area of the garden is 196 square meters, what are its dimensions? Thank You
Click here to see answer by RAY100(1637) |
Question 193790: a retailer spent $48 to purchase a number of special mugs. Two of them were broken in the store, but by selling each of the remaining mugs for $3 above the original cost per mug, she made a total profitt of $22. If the price for the n mugs is $48, how can we express the cost per mug? Now have to write in standard form of a quadratic equation
Click here to see answer by solver91311(16897)  |
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