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Thread: Radicals and Quadratics kick my arse

  1. #1
    Sane Scientist Bahamut2000X's Avatar
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    Default Radicals and Quadratics kick my arse

    So I have an exam coming up in my Algebra class for college. Unfortunately I've done terrible on the last test and need to do good to pass this class. In fact I'm looking at having to score at least 158 points out of 200 to pass on this exam.

    Unfortunately for me I'm terrible at doing Quadratics and Radicals. My teacher is an insane mathematician and thus tries to teach like we all were in it to become math majors also. Which is terrible to teach those just trying to pass, and even more so for me as I don't learn in that way at all. I never knew you could spend close to an hour ON A SINGLE PROBLEM!

    Regardless I'm working on studying for this final but I suck at studying by myself, so I'll ask for any of the math people here on the forums to give me some tips and help on doing these problems as I suck quite a lot at doing them by myself as I've learned.
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    One Hundred Chimneys Recognized Member Tavrobel's Avatar
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    Radicals are fun. And quadratics are fun, too. What specifically did you have in mine when you decided to make this thread? There's a lot of stuff to cover.

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    Sane Scientist Bahamut2000X's Avatar
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    Haha indeed I should probably clarify so I don't go over the wrong things.

    Well as far as quadratics go, I need to know how to complete using the square, or at least on the last test I had to get to the third step of figuring it out. Also using the quadratic equation to solve, algebraically that is, so I don't have an exact answer but like a radical or some such nonsense.

    As far as radicals my biggest issue is I can't understand how you simplify it half the time. Such as where you would end up with something random like a 2 radical 7 (just to make up something random) I understand the basic idea that you would split the first number under the radical into a perfect square and what not, but sometimes I see something where I don't see a perfect square and I still see it being simplified.

    That's all off the top of my head I can think of at the moment.

    Although on a related note my graphing calculator has a major problem where it won't graph. Every time I try to graph something, even as simple as a line as Y = 3 I always get a "Error: Invalid Dim". I've been playing around with it, but I have yet to figure out why it would do this.

    Anyways I appreciate the help.
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    One Hundred Chimneys Recognized Member Tavrobel's Avatar
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    CTS is really weird, but it saves your life, especially in the higher levels.

    So let's say your function is (X<sup>2</sup> + 18X + 7). X is squared and without a coefficient, so, no worries there. However, the 18X is the biggest indicator of where to start. Since we want a number that when doubled equals 18, divide it by 2. You'll get 9. If your function were (X<sup>2</sup> + 18X + 81), then your life would be much easier. 81 - 7 = 74, so now we add 0 to the equation in the form of (+74 - 74). The function now looks like (X<sup>2</sup> + 18X + 7 + 74 - 74). Combine like terms. (X<sup>2</sup> + 18X + 81 - 74). Move the -74 to the side, and the first three terms are in the form of a polynomial square. (X<sup>2</sup> + 18X + 81) --> (X + 9)<sup>2</sup> - 74. Do whatever else you need from there. Set it equal to something, integrate, derivate, whatever is applicable.

    A general rule of math is that sometimes answers aren't pretty. For example, Radical(28) --> 2R(7). It's not pretty, but that's what it is. If you must when you see a big number, break it down into its prime factors. If you see something twice, take it out.

    Make sure your graph isn't set to highlight one of the Plot# at the top. To get the highlight off, just hit up and hit the Plot# again.

  5. #5
    Draw the Drapes Recognized Member rubah's Avatar
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    or you know, to put it simply:

    "you've got a quadratic: x squared, x, constant, all those added together equal 0. Put the constant on the other side of the eqation, take half the x, square that, add it to both sides, solve for the two xes!"

    and uh, so I say something that tav hasn't (even if more simply), when factoring, you can see if something's divisible by three by adding together literally each number that composes a number. for example, 539 = 5+3+9 = 17. 17 isn't divisible by three (1+7 = 8) so 539 is not divisible by three. (but 651 is!)

  6. #6
    I'm selling these fine leather jackets Aerith's Knight's Avatar
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    one word: abc-equation

    ..

    okay two words.

    ax^2 + bx + c = 0

    D = b^2 - 4ac

    x = (b +/- sqrt(D)) / 2a


    Works every time.

    for the rest of them there are simple steps such as

    x^8 = (ax+1)^8

    x = ax + 1

    ..

    For logs always remember

    10 log 1000 = 3

    so 10^3 = 1000

    And substitute x for something in that and you know the solution.

    If you need a radical just use the differential of it.

    y(x) = ax^2 + bx + c

    dy/dx = 2ax + b

    Use dy/dx = 0 if you want to know the optimum or minimum.

    Intergrate if you want to know the surface beneath the graph. The total amount.

    y(x) = ax + b

    F(x) = 1/2 ax^2 + bx + C

    For surface of a specific point use

    F(x1) - F(x2)

    You would get the surface between x1 and x2
    Last edited by Aerith's Knight; 06-13-2008 at 07:55 AM.


  7. #7
    GO! use leech seed! qwertysaur's Avatar
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    Quote Originally Posted by Aerith's Knight View Post
    one word: abc-equation

    ..

    okay two words.

    ax^2 + bx + c = 0

    D = b^2 - 4ac

    x = (b +/- sqrt(D)) / 2a


    Works every time.
    If you copy it down correctly. It should be

    x = (-b+/-sqrt(b<sup>2</sup>-4ac))/2a

    Gotta love the quadratic formula. It will always work. But make sure you are in standard form (ax<sup>2</sup>+bx+c=0) before completing the square, using synthetic division, factoring or use the quadratic formula.

  8. #8
    Sane Scientist Bahamut2000X's Avatar
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    Thanks for the help but I should probably mention my finals have been over since Monday. Though I (probably) failed. Still waiting on the grades to get posted.

    Oh and ends up there was like 3 radical questions, and 2 quadratics, everything else on the test was EVERYTHING else I didn't study for but knew slightly better.
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  9. #9
    I'm selling these fine leather jackets Aerith's Knight's Avatar
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    Quote Originally Posted by qwertyxsora View Post
    Quote Originally Posted by Aerith's Knight View Post
    one word: abc-equation

    ..

    okay two words.

    ax^2 + bx + c = 0

    D = b^2 - 4ac

    x = (b +/- sqrt(D)) / 2a


    Works every time.
    If you copy it down correctly. It should be

    x = (-b+/-sqrt(b<SUP>2</SUP>-4ac))/2a

    Gotta love the quadratic formula. It will always work. But make sure you are in standard form (ax<SUP>2</SUP>+bx+c=0) before completing the square, using synthetic division, factoring or use the quadratic formula.
    I see I forgot a minus sign there..

    Sorry about the test bahamut.


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