Hello, I am Nerdus Pegasus (AKA Extreme Magnetic Power), and I LOVE math. Why did I name myself "Nerdus"?. Straight 'A's and stuff. Yeah. In school, I see so many people around me struggle with math, and I just want to help them out. So, I've started this thread so that I can help you with math. Rules and disclaimers: Questions can be about anything math. This includes discussion on math, simply asking a math question, or asking about the usefulness of math, like in a job or daily life. Please provide an example for all of your questions. You may post many questions at a time, but please provide an example for all of them. If there is a picture with extra information on it, please try to take a picture, draw it, describe it, or make an ASCII art of it. Please keep math jokes to a minimum; for example: sin(gerine) / cos(gerine) = tan(gerine). My help is limited to my knowledge. I can help you in basic math, algebra, geometry, and trigonometry (which I am learning in school right now). Don't expect me to know multi-variable calculus. I will attempt to answer your questions, but if I am unable to help, I will respond that I can't help. Don't expect me to give you the answer. I want to help you expand in your knowledge of math. I will only help you find the answer, perhaps with a different example. I may even create my own example. Please ask your questions in advance. I don't want you to ask a question the day before the test. Anyone else proficient at math may help. Please make sure that you give thorough explanation, as you are the teacher. FAQs, guidelines, conventions, and other helpful stuff: When using exponents, there is a [SUP]superscript[/SUP] button for that. If you need to do a function, like sine or square root, use parentheses to show what is inside the function: sin(x), sqrt(546) = 23.3666428911. Operators will be presented in a syntax similar to the C programming language. = means equals; >= means greater than or equal to; <= means less than or equal to, > means greater than, < means less than, != means NOT equals, =/= also means NOT equals, * means multiply, and % means modular division. I am from the United States of Awesomeness. I say math, not maths. I don't care which way you say it, though. Use common sense! Add stuff that may be helpful, even if it isn't stated here. More FAQs may be added if needed. Anyways, I hope I can help!
I SO could have used you the other day! I just failed a math test today on long division and the quadratic formula. Like... find all zeroes of 24x[SUP]5[/SUP] + 7x[SUP]3[/SUP] + 14x[SUP]2[/SUP] + x + 13 ...all the long division... all of it... ;A;
http://wolframalpha.com - all the help you'd ever need. Also this may be good to move to the Community Help forum
This one is very hard to simplify. Graphing it, the only real zero I get is about (-.9512329, 0) Are you sure it wasn't an easier problem? For example, 24x[SUP]3[/SUP] + 8x[SUP]2[/SUP] + 12x + 6 = 0 For this one, I would solve by grouping. First I put the equations in parentheses to visualize it easier. (24x[SUP]3[/SUP] + 8x[SUP]2[/SUP]) + (12x + 4) = 0 Then, I factor the x's and other factors. 8x[SUP]2[/SUP](3x + 1) + 4(3x + 1) = 0 Then, I factor out the (3x+1). (3x+1) * (8x[SUP]2[/SUP] + 4) = 0 Then, I solve using the zero product property. 3x + 1 = 0 3x = -1 x = -1/3 8x[SUP]2[/SUP] + 4 = 0 2x[SUP]2[/SUP] + 1 = 0 2x[SUP]2[/SUP] = -1 x[SUP]2[/SUP] = -1/2 x = (plus-minus)sqrt(-1/2) negative square roots are not real. My answer is x = -1/3 I would prefer having questions that were actually on the test, other than made up ones witch are way harder (unless if your question was an actual question). - - Auto Merge - - Good point. Please move me. All hail the great Wolfram Alpha... All hail the great Wolfram Alpha... All hail the great Wolfram Alpha...
Yes, it was probably easier. I was throwing random numbers into an equation (and skipped 4 because the damn test skipped it too). What you missed was that we were working with imaginary. I think one of the zeroes to the equation I don't remember was 3i. It also has the [SUB]5[/SUB] exponent, so there has to be, like, 5 or so answers.
u... wut Dat brazenly obvious division by zero on line 4 renders that false, as it is not a legitimate operation. To elaborate: At the fourth step, you have (a+b)(a-b) = b(a-b) Intuitively, you'd divide to get that b on the right by itself. BUT WAIT a = b Thus, (a-b) = 0, so you'd find yourself with two (a-b)/(a-b) parts. And zero over zero is indeterminate form. As in, it cannot be assigned a specific value for the simple reason of anything multiplied by zero gives you zero. 0/0 = ? is asking 0 x ? = 0, which can literally be assigned any value and grant the same outcome. I'm explaining this because of its difference with flat out division by zero; one is undefined, the other is indeterminate. So yey, division by zero is a no-no. I've seen this one and discussed it with my teacher before.
If you can remember an actual question on the test, then maybe I can actually help. You can help people that ask about calculus, if you wish. It's perfectly optional for you to help. Zatchiel is right. You cannot divide 0 by 0. This is what is called an indeterminate form. Other indeterminate forms include (but are not limited to): 0[SUP]0[/SUP], 1[SUP]∞[/SUP], ∞ - ∞, and 0 * ∞. Some undefined forms include (but are not limited to): 1/0 and ∞/0. Indeterminate forms, in short, cause math to break if not treated properly. These should only be examined in a calculus class. Further reading on indeterminate form: http://en.wikipedia.org/wiki/Indeterminate_form Here is a Numberphile video on dividing by zero: http://www.youtube.com/watch?v=BRRolKTlF6Q
Quadratic Equations my good sir. I get how to do it on a very basic, very time consuming level, though it's the more advanced methods that baffle me.
Well, to factor any x[SUP]2[/SUP] + bx + c expression, I must first determine what 2 factors of c (let's call them n and m) can be added together to get b. We can factor c into it's prime factors in order to determine what 2 factors can go into c. For example, take x[SUP]2[/SUP] + 13x + 36. 36 has the prime factors 2, 2, 3, and 3. In order to get 13, I can multiply 2*2 to get 4, and 3*3 to get 9. 4 + 9 = 13. So, the factorization of x[SUP]2[/SUP] + 13x + 36 is (x + 4)*(x + 9). I'm not sure what you mean by "advanced methods."
By "perfect squares," I think you mean solving by completing the square. Perfect square quadrilaterals are always in this form: x[SUP]2[/SUP] + bx + (b/2)[SUP]2[/SUP] These can be factored like so: (x + b/2)[SUP]2[/SUP] I will now demonstrate how to solve by completing the square. x[SUP]2[/SUP] + 8x + 15 = 0. Move the 15 to the other side by subtracting on both sides. x[SUP]2[/SUP] + 8x = -15. Knowing that b = 8, I find that b/2 = 4. 4[SUP]2[/SUP] = 16, so I shall add 16 to both sides. x[SUP]2[/SUP] + 8x + 16= -15 + 16. x[SUP]2[/SUP] + 8x + 16= 1. x[SUP]2[/SUP] + 8x + 16 is (x + 4)[SUP]2[/SUP], so (x + 4)[SUP]2[/SUP] = 1 x + 4 = (plus-minus) 1 x + 4 = 1 x = -1 x + 4 = -1 x = -5 So, x = -1,-5 Grouping: Solving by grouping is for cubic equations. x[SUP]3[/SUP] + 3x[SUP]2[/SUP] - 4x - 12 First, I shall order the front and back of the equation into groups. (x[SUP]3[/SUP] + 3x[SUP]2[/SUP]) + (-4x - 12) (Note that I took that minus in the center and put it with the -4x.) Next, factor out the common factors in each group. x[SUP]2[/SUP](x + 3) + -4 (x + 3) Next, factor out the (x + 3). (x + 3)(x[SUP]2[/SUP] - 4) We are done with grouping. Lastly, factor the (x[SUP]2[/SUP] - 4). (x + 3)(x + 2)(x - 2) Hope this makes more sense! I need to go to bed now, so I won't be able to respond until tomorrow. - - Auto Merge - - I try and explain as clearly as possible! If I need to explain anything again, just ask.
Ah there we go. Grouping was what usually made my head spin, it's a bit tricky to remember after a few semesters. Thank you very much for the help, it's very appreciated.
I suck at math but i'm not taking the Compass test to get into College until the Fall so I guess i'm fine for now.
