12172002, 06:52 PM  #1 
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Help please... (clarified some)
Okay long story short I am suffering from a chemical inbalance and dealing with my body adjusting to 8 different medications. Unfortunately I am in the middle of a school semester and some of the things I have been dealing with are confusion, lack of concentration, and basically shutting down when I have frustration caused by the previously 2 mentioned symptoms.
Anyway, I have 2 chapters left in math and while I hesitate to ask for help, I have 2 days to do 2 chapters or I get no credit for the course. Here are 3 problems that I have scanned in hopes that someone can help me. http://www.sixanswers.com/esc/images/math.jpg I know that helping isn't the easiest thing when computers aren't great at typing math problems. If anyone can help I will assume that x^2 is x with an exponential of 2 or "x to the 2nd". Thanks. I know.. this isn't hard math Last edited by cnjmorris; 12172002 at 09:04 PM. 
12172002, 07:07 PM  #2 
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If this is what youre looking for then yay, if not then disregard.
10. x = b +/ square root( b^2  4(a)(c))  2a 10 +/ square root (100 4(2)(7))  4 10 +/ 6.63 (rounded to two decimal places)  4 Think that should do it. Good luck EDIT: this problem is number 10, not 11. And the two answers for this are the x intercepts for the parabola that will be graphed. Last edited by rayndor1.0; 12172002 at 08:03 PM. 
12172002, 07:18 PM  #3 
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Help please...
I appologize I should have specified I need to graph #10 and #11 on a plane. I can figure out the graphing part if I can just figure out how to do the work to arive at the answers.
#12 I know the answer is (0,3) , (4,0) I just don't know how to explain the work. Sorry for not being more clear, and thanks. 
12172002, 07:38 PM  #4 
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If you are supposed to graph them, for number 10 you will just have two points on a number line, go through what rayndor said and you will have x=4.679 and x=.321. The only way to graph an equation in one variable is on a numberline, not a plane.
for the second one, you have x=4^2+4y so you square the 4 to get x=8+4y then you ad 8 to both sides giving you x+8=4y divide both sides by 4 and you get (1/4)x+2=y or y=(x/4)+2. Enter that into your calculator or make a table and plot it by hand (which ever way you are supposed to) On the last one are you supposed to find where those two lines intersect? Last edited by Igota; 12172002 at 08:16 PM. 
12172002, 08:09 PM  #5 
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More stuff.
For #11, use the same process as was used for #10 (quadratic equation), except that the answers will be the y intercepts and the parabola will be rotated 90 degrees. For 12, im not sure if your teacher wants you to substitute or what, but an explanation for getting the answer would be when setting y=0, x is 4. When x=0, y is 3. I know there is another way but i cant remember it at this point and time. Last edited by rayndor1.0; 12172002 at 08:15 PM. 
12172002, 08:41 PM  #6 
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If you are allowed to graph them on a calculator and you are supposed to find the intersections of the graphs for #12, solve the first equation for y and you will get a two part solution y=(3*squareroot(16x^2))/2 and y=(3*squareroot(16x^2))/2 graph both those equations and it will give you the picture of the first equation (which will be a squashed circle (elipse) crossing the xaxis at +4 and 4 and the yaxis at +3 and 3) the second equation you will solve for y as well, giving y=(3/4)*x+3 which is a line that with slope negative 3/4 and prossing the yaxis at +3 so one point of intersection is (0,3), the other point is (4,0), since slope is rise(y)/run(x) you will go from (0,3) negative 3 y and 4 x, leaving you at (4,0).

12172002, 08:57 PM  #7 
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Help please...
Number 10 this is the graphed answer:
http://www.sixanswers.com/esc/images/10.jpg y=2x^210x+7 x=b +/ sqr(b2  4(a)(c)) 2(a) x=10 +/ sqr(100  4(2)(7)) 2(2) x=2(5 +/ sqr(11)) 2(2) x=5 +/ sqr(11) 2 I assume that I have done okay up till there.. but how I get from there to a graph is the hard part. Number 11 this is the graphed answer: http://www.sixanswers.com/esc/images/11.jpg This one I am a bit confused on. Its "y^2" not "4^2", I may have written it poorly in the original problem. x=y^2+4y Do I "complete the square" by (b/2)^2 or (4/2)^2 ? x+4 = y^2 +4y +4 Even if that is the right path, I have no clue where to go from there. I have been doing this crap for weeks in different variation but now I am drawing a blank and none of the examples I have are in similar form. 
12172002, 09:05 PM  #8 
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ax^2 + bx + c is the original equation. The two answers you recieved from using the quadratic equation can each be plugged into the original equation to prove they are correct.
In regards to the graph, those same two answers are the intercepts on the graph, and c is where x becomes 0 (the parabola crosses the y axis. Im not real sure where you are getting lost as you seem to have it figured out in the layout you presented. EDIT: the "b" at the beginning is having a 10 plugged into it so the negatives cancel and it is just 10. That may clear up some confusion 
12172002, 09:17 PM  #9 
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Sorry for the misreading of the equation before, anytime you graph a parabola you want to find the vertex , to do that you do the old "completing the square" which puts a parabola into standard form : y = a(x  b)^2 + c, where a, b, and c are constants. Once you have the equation in standard form, the vertex is at the point (b, c) and the line of symmetry is x = b. If a is positive, the parabola opens up, if a is negative, the parabola opens down.
That is the general way to do it (will work for problem #10), now for what you need on #11. Since the original equation is x = ay^2 + by + c, then it is a parabola which lies "on its side" and opens either to the left or the right (left in your case cause a is negative). You can use completing the square to put the equation in standard form, x = c(y  h)^2 + k, with its vertex at (k, h) and line of symmetry y = h. To plot the parabola, I usually find that it is easiest to create a table of values using a few points on either side of the vertex to find some points which lie on the parabola. 
12172002, 09:50 PM  #10 
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Help Please....
Thanks for the help so far. I am beginning to remember some of what I have been working on and its making sense. The funny thing is I kind of thought that was how to do #10 but it seemed too hard compared to the problems I had been doing. This book has a bad habit of jumping around in its teaching and assuming a lot, never giving examples even remotely similar to the questions it asks. I know that in math that synthesis is important, but so is good old fashion teaching.
Anyway here is the part where I am confused. I tried completing the square and ended up with x=(y^2 +4y+4) 4 but the part that confuses me is the negative y. Sure I see that it should factor to x=(y+2) 4 but the negative throws that off. Thanks again. 
12172002, 10:09 PM  #11 
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take the equation x=y^2+4y and multiply both sides by 1 resulting in x=y^24y complete the square by adding 4 to the right side (and make sure you add 4 to the left side as well to ballance out the change) you now have x+4=y^24y+4 factor the right side and you get x+4=(y2)^2 subtract 4 from each side gives x=(y2)^24 and multiply both sides by 1 again to get the final equation x=1(y2)^2+4
a=1 so it opens left, h=2 and k=4 so the vertex is (4,2), then plot some points. let x=0 and y will be 0 or 4, at x=2 y=3.4 and .6, at x=3 y=3 or 1, and at x=2 y=4.4 or .5 Last edited by Igota; 12172002 at 10:12 PM. 
12182002, 05:04 AM  #12 
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Not doubting your story bro but wow! With a response like this, I wish I'd had these boards when I was in school. 'sploit, sploit!!
More on topic, Igota's solution appears correct to me
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12182002, 08:15 AM  #13 
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Progress...
Heh, thanks for not doubting the sotry, I wish I was in a position to lie about it.
I knew I could count on the people here to help, they have helped me on a few problems in the past. I actually hold off and work hard to figure out the stuff.... but after spending most of yesterday working on those 3 I realized it was now or never. There may be a few people here who would answer every question I posted just to show off how smart they are, not sure... but I do know that the people who have helped thus far most likely wouldn't help if someone posted their whole math assignment. I asked for 3 out of my 52. Of course I am only half way done, so hoping I don't need more help =). Notice I didn't ask for answers though, I just wanted to know how to do the work. I actually enjoy math, I like the feeling I get when it finally clicks, but I too have been out of school for 10 years trying to go back straight to this =) I do agree though, that this forum is a blessing, since I have been able to ask about math, computers, speakers, etc, etc. Lets face it, their is a wide variety of experience available here. Just wish I knew enough about more topics to answer others questions when asked =) 
12182002, 12:38 PM  #14 
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I tutored when I was in college, and I'd say that the hardest working, most earnest students I had were those that were coming back to higher education after years in the workforce (even only being out of school a few years I'm amazed at how much effort it takes to remember the stuff I don't use in my day to day).
My point is I respect your going back to it, especially in light of your medical status (I don't pretend to know anything about it, but I barely function with a head cold, let alone a true chemical imbalance :P )
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12182002, 08:10 PM  #15 
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YOU BASTARDS!!
Here I was, living life in relative bliss and then I had to check this thread out and be confronted with the living nightmare that I faced for three years in electronic engineering!! *shudder* ohh the horror! /faint 
12182002, 11:34 PM  #16 
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dude, that was like ninth grade. not really brain surgery.

12192002, 07:49 AM  #17 
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Cnjmorris...
Check out the disabilities resource center in your college. There are programs to assist you in studying and taking tests... also let your professors know of your chemical imbalance. The should be able to accomdate that in test taking, etc. If not, contact the provost office in your school
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12192002, 11:16 AM  #18 
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Diabetes/chemical help...
Well, to some degree I am lucky, because I am taking a lot of courses via the internet and correspondence meaning I have no actual timed tests.
The problem is that it means that I have less resources at my disposal, such as actual physical teachers. I can do work on my schedule but if I shut down and get behind it is only hard to catch up. I don't have time right now to look into whats available and I am hoping that by next semester a lot of these problems will be sorted out. Thanks for the input though. I may look into it later just in case I need it in the future. 
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