Tutors Answer Your Questions about Quadratic Equations (FREE)
Question 668261: I am trying to understand if the equation I have done is anywhere near correct. I think that I understand steps 1-4 correctly but the last two I do not which then makes my answer incorrect. I have two more problems to do like this one; if I cannot understand the first one then the rest will not be correct. Here is what I have so far.
Problem A: x2 – 2x – 13 = 0
Step 1: I need to move the constant term which is “13” to the right side too. As always we need to remember that what we do to one side we have to do to the other.
X2 – 2x – 13 = 0
X2 – 2x – 13 + 13 = 0 + 13
X2 – 2x + 0 = 0 + 13
X2 – 2x = 0 +13
X2 – 2x = 13
Step 2: Here I am going to multiply each term by four times the coefficient of the x term.
4 * 1 =
4x2 – 8 x = 52
Step 3: Square the coefficient of the original x term and add it to both sides of the equation. Throughout my equation I have learned that the x term is 2. However, for this step I need to square 2 and when I do that equals 4. I will add 4 to both sides of my equation like this.
4x2 – 8x + 4 = 52 + 4
4x2 – 8x + 4 + = 56
Step 4: Take the square roof of both sides.
(2x-2) * (2x-2) = 4x2 - 4x - 4x + 4
4x2 - 8x + 4 the expression on the left side of the equation
(2x - 22) = 56
2x - 2 = +/- sqrt(56)
Step 5: Set the left side of the equation equal to the positive square root of the number on the right side and solve for x. Once again remember what you do to one side you will have to do to the other.
2x - 2 = sqrt(56)
2x = sqrt(56) + 2
At this time I will need to divide both sides of the equation by 2 to and once I do it will look like this.
x = sqrt(56) + 2) / 2
Step 6: Set the left side of the equation equal to the negative square root of the number on the
right side of the equation and solve for x.
2x - 2 = - sqrt(56)
2x = - sqrt(56) + 2
divide both sides of the equation by 2 to get:
x = sqrt(56) + 2) / 2
Click here to see answer by ReadingBoosters(3246)  |
Question 668261: I am trying to understand if the equation I have done is anywhere near correct. I think that I understand steps 1-4 correctly but the last two I do not which then makes my answer incorrect. I have two more problems to do like this one; if I cannot understand the first one then the rest will not be correct. Here is what I have so far.
Problem A: x2 – 2x – 13 = 0
Step 1: I need to move the constant term which is “13” to the right side too. As always we need to remember that what we do to one side we have to do to the other.
X2 – 2x – 13 = 0
X2 – 2x – 13 + 13 = 0 + 13
X2 – 2x + 0 = 0 + 13
X2 – 2x = 0 +13
X2 – 2x = 13
Step 2: Here I am going to multiply each term by four times the coefficient of the x term.
4 * 1 =
4x2 – 8 x = 52
Step 3: Square the coefficient of the original x term and add it to both sides of the equation. Throughout my equation I have learned that the x term is 2. However, for this step I need to square 2 and when I do that equals 4. I will add 4 to both sides of my equation like this.
4x2 – 8x + 4 = 52 + 4
4x2 – 8x + 4 + = 56
Step 4: Take the square roof of both sides.
(2x-2) * (2x-2) = 4x2 - 4x - 4x + 4
4x2 - 8x + 4 the expression on the left side of the equation
(2x - 22) = 56
2x - 2 = +/- sqrt(56)
Step 5: Set the left side of the equation equal to the positive square root of the number on the right side and solve for x. Once again remember what you do to one side you will have to do to the other.
2x - 2 = sqrt(56)
2x = sqrt(56) + 2
At this time I will need to divide both sides of the equation by 2 to and once I do it will look like this.
x = sqrt(56) + 2) / 2
Step 6: Set the left side of the equation equal to the negative square root of the number on the
right side of the equation and solve for x.
2x - 2 = - sqrt(56)
2x = - sqrt(56) + 2
divide both sides of the equation by 2 to get:
x = sqrt(56) + 2) / 2
Click here to see answer by MathTherapy(10551)  |
Question 669615: I am so confused on this.
Graph each quadratic function, and state it's domain and range.
y=-2x^2+3
The directions are as follows:
•There are many ways to go about solving math problems. For this assignment you will be required to do some work that will not be included in the discussion. First, you need to graph your functions so you can clearly describe the graphs in your discussion. Your graph itself is not required in your post, although the discussion of the graph is required. Make sure you have at least five points for each equation to graph. Show all math work for finding the points.
•Specifically mention any key points on the graphs, including intercepts, vertex, or start/end points. (Points with decimal values need not be listed, as they might be found in a square root function. Stick to integer value points.)
•Discuss the general shape and location of each of your graphs.
•State the domain and range for each of your equations. Write them in interval notation.
•State whether each of the equations is a function or not giving your reasons for the answer.
•Select one of your graphs and assume it has been shifted three units upward and four units to the left. Discuss how this transformation affects the equation by rewriting the equation to incorporate those numbers.
•Incorporate the following five math vocabulary words into your discussion. Use bold font to emphasize the words in your writing (Do not write definitions for the words; use them appropriately in sentences describing the thought behind your math work.):
◦Function
◦Relation
◦Domain
◦Range
◦Transformation
3
Click here to see answer by ewatrrr(24785)  |
Question 669481: a man travels a total distance of 166km/hr. his average speed for the first 130km is xkm/hr and for the remaining journey , (x-25)km/hr. if the whole journey takes 3 hours and 15 minutes , form an equation in terms of x.
solve this equation and hence find the time he would have saved if he maintains an average speed of xkm/hr throughout ?
Click here to see answer by ankor@dixie-net.com(22740)  |
Question 669596: I need help with this problem of algebra 2:
The general form of an equation for a parabola is y=ax^2+bx+c, where (x,y) is a point on the parabola. If three points on a parabola were (2,-10)(-5,-101)and (6,-90) determine the values of a, b and c write the general form of the equation.
Click here to see answer by ankor@dixie-net.com(22740) |
Question 670740: Which of the following statements describes one part of completing the square for x² + 4x = 32?
a)Take the square root of 36 and add 2.
b)Take the square root of 32 and subtract 2.
c)Take the square root of 36 and subtract 2.
Click here to see answer by checkley79(3341) |
Question 671426: A place kicker kicks the ball upward with a velocity of 32ft/sec. Ignoring the height of the kicking tee, how long after the ball is kicked does it hit the ground? Use the formula h(t)= -16ft squared + vt where h(t) is the height of the object in feet, v is the ball's initial velocity in ft/sec, and t is time in seconds. What is the highest point reached by the ball?
Click here to see answer by Alan3354(69443)  |
Question 672423: Answer the questions below about the quadratic function.
f(x)= 3x^2 +24x + 51
Does the function have a minimum or maximum value?
What is the functions of the minimum or maximum value?
Where does the minimum or maximum value occur? x= ?
Click here to see answer by stanbon(75887) |
Question 672615: A bicyclist is riding at a speed of 18mi/h when she starts down a long hill. The distance d she travels in feet can be modeled by d(t) = 4t^2 + 18t, where t is the time in seconds. How long will it take her to reach the bottom of a 400-foot-long hill?
We have to use the quadratic formula to find t.
I attempted the problem and i got 12.5 seconds.. but im not sure if im right
Heres my work:
d(t)=4t^2+18t=400
d(t)=4t^2+18t=400/4
t^2+4.5t=100
t^2+4.5t-100=0
(x-12.5)(x+8)
12.5 & -8 seconds.
is this correct?
Thanks(:
Click here to see answer by jim_thompson5910(35256) |
Question 672785: A soccer ball is kicked in the air the height h(t) of the ball in metres according to time in seconds is expressed as a function h(t) = -2t^2+8t. For what interval of t does the height of the ball increase??
Click here to see answer by stanbon(75887) |
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