Print

Please login or register.
1 Hour 1 Day 1 Week 1 Month Forever
Login with username, password and session length

[Culise]

4x^-12x+9=0

Show me the next step from there, if you can please.

Just to confirm, do you mean 4x² or 4x^-12x?  Because the two are very different, and the latter is not something you'd expect to see in beginner algebra where you're just starting factorization.  I'll assume the former for now, but you should bear the notation in mind since you seem to be consistently using just the power notation symbol ^ without actually specifying the power (It'd be like saying 2+= 4, which I've not seen since my old C++ class).  ^_^

Right, second-order factorization for beginners.  Now, I'm no good at explaining things, either, but there are a few ways to go about this.  You have your equation 4x²-12x+9.  Now, you need to figure out how to break it apart (we'll assume you're already told that you need to break it apart).  We'll go about this the other way: We know that there is some (Ax+B)(Cx+D) that'll spit out what we want: (Ax+B) and (Cx+D) are the "factors" involved that can be multiplied together using the FOIL rule-of-thumb Descan wrote out.  Just for reference, we'll write down those equations for this particular problem, just to be complete.

• A*C = 4
  
• B*D = 3
  
• B*C+D*A = 12

Nice, but not nice; we have a system of three equations for four unknowns.  At any rate, let's look at the coefficients of the full equation: 4, -12, and 9.  These are what we'll be breaking down, so let's see how we can do so:

• 4 is either 2*2 or 1*4, or negative permutations therein: -2*-2 or -1*-4
• -12 is either -1*12, 1*-12, -2*6, 2*-6, -3*4, or 3*-4
• All of these in turn breaks down to some variation on -2*(2*3), 2*(-2*3), or 2*(2*-3), depending on where the negative-sign ends up
• 9 is either 1*9 or 3*3, or negative permutations therein: -1*-9 or -3*-3

You'll notice some common factors here - specifically, 2, 4, and 3.  In particular, neglecting the signs, there are only two possibilities for A*C and B*D in our factors, so the real sticking point is how to put them together to get 12.  Well, did you notice the simplification I indented in the above, under 12?  If A and C were the same (that is, if we went with 4 = 2*2, making both A and C 2) and B and D were the same (likewise, with 3), why, you'd end up with 2*(B*C) or 2*(A*D), which are the same: 2*(2*3) = 12; we're almost there, but we missed the sign!  That means that there must be a negative sign somewhere in there: either A and C are -2, or B and D are -3, but it can't be both, because they'd cancel.  Let's try each:

• 2*(-2*3)=-12
• 2*(2*-3)=-12

...well, that's interesting; they both work.  And that means that you have not one, but two ways to factor this problem.  The first is the factorization Karlito gave, and the second is the following:

(-2x+3)(-2x+3)=(-2x+3)²

This actually happens because it's the same factor, whether you go with (2x-3) or (-2x+3)=-(2x-3), squared.  Specifically, it's the same reason why 4 is both 2*2 and -2*-2 - the negative signs will cancel out when you multiply out, which can be shown by simplifying one of them:

(-2x+3)² = [(-1)*(2x-3)]² = (-1)² * (2x-3)² = 1 * (2x-3)² = (2x-3)²

Two different ways to factor the same equation, and you can actually pick either.  Because they're the same, they'll both give you the same value for x in the end when you solve the full formula. 

There are more complex explicit methods that can be used to factor polynomials (Kronecker, I do not miss you at all), and I could try to go into them if you wish, but it's been quite some time since I've done it.  A lot of the above is just rule-of-thumb and practice, along with educated guessing (for instance, spotting that the coefficients are all 2s or 3s). 

EDIT: I really wish there was an [indent] tag, though.  It'd make those equations much easier to pick out.  Formatting tweaks and some wording adjusted.

[ShadowHammer]

Culise's solution is accurate, but here is another way of doing it that I found easier when I first learnt to factor (4 months ago )
4x²-12x+9

     Find two numbers that add to make -12x. The first should have a factor in common with 4x², and the second should have a factor in common with 9x.
          The first number is -6x, because it, like 4x², can be divided by 2x.
          The second number is also -6x, because it, like 9, can be divided by -3.
          Both numbers added together (-6x+(-6x)) make -12x.
     Take these two numbers and separate -12x into them.
=4x² -6x -6x +9
     
     Add brackets to make things easier.
          Note how the - is moved inside the bracket along with the 6x, leaving a + outside the brackets.
=(4x² -6x)+(-6x+9)

     Take the common factor out of each bracket
          Remember from step 1; the common factor of 4x² and -6x is two, while the common factor of -6x and 9 is -3.
               When you "take out" a common factor, divide the contents of the bracket by that number and then multiply the entire bracket by it; this rearranges the equation   
               without changing the value of the bracket.
=2x(2x-3)-3(2x-3)

     Now, take the two things outside the brackets, 2x and -3, and put them together in their own bracket. The other two brackets should be the same; if they aren't, you did something wrong. Take the contents of those two brackets and make it one bracket.
=2x(2x-3)-3(2x-3)
=(2x-3)(2x-3)

     Now, since both brackets are the same, you can write it as a square, because a square is a number multiplied by itself.
=(2x-3)²
I seem to have difficulty phrasing step 1; if someone else could rephrase it, that would probably be helpful. It is pretty much the same as Culise's

4 is either 2*2 or 1*4, or negative permutations therein: -2*-2 or -1*-4
-12 is either -1*12, 1*-12, -2*6, 2*-6, -3*4, or 3*-4
-snip-
9 is either 1*9 or 3*3, or negative permutations therein: -1*-9 or -3*-3

Actually, looking over it again, this method might be the same as Culise's, just explained differently. Oh well, hope it helped.

[vagel7]

So guys, what's the best method to learn mathematics and to grasp the subjects that you need at the time? And I don't just mean the theory part, but actually the solving problems part.

Math is actually a bit difficult in uni

[Helgoland]

What sort of mathematics?

[wierd]

Exactly what kind of problems do you have when doing math?  Is there a specific kind of problem that always busts your nuts?  It might be that there is something that just never "cklicked" before that could have been better explained in primary school, that is holding you back.

Can you give an example problem that you are having issue with?

[vagel7]

Well, for some reason I am having problems with algebra and calculus, while some more "abstract" categories of math like discrete maths is going pretty well for me. Also the problem is not so much about not knowing the theory, as in applying that theory into practice.

[Vector]

.

[vagel7]

I wouldn't call problems, that have multiple square roots and absolute values in them to be exactly one-step. One-step is something that we do in primary school.

[wierd]

Not one-step, no---  but just break the problem down into steps.

Absolute values are easy.  Square roots can be a pain, but if you know your properties, sometimes you can negate the square rooting using the distributive and commutative properties.

At this point in math, they are trying to get you to understand and use the properties of the numbers, not the basic arithmetic. They presume you already know how to add, subtract, multiply, divide, exponentiate and factor. Instead, they want you to understand how you can exploit the properties of numbers to accomplish things.

[Bauglir]

Step one: Replace all roots with fractional powers. Square root, for instance, is "to the 1/2th power". Now you can use all the same rules you normally do for working with exponents. Why they even teach radical notation I don't know (maybe somebody who knows things can explain >__________>).

Anyway, for algebra the main skill, in my experience, to work on is grouping things together. Minimizing the number of constants floating around is a good goal to work toward.

Calculus is similar, but there's less concrete a goal here. As Vector said, though, your strategy where you're at will generally be to hammer what you're given into a format that suits a general rule you already know. Once you've got your table of simple derivatives and integrals, the bulk of the techniques they give you are for doing that transformation into components you already know. If you've got the theory worked out, it becomes a matter of mapping the theory to the techniques you're given so that you understand why they do the things you're told they do (and that's such a deeply internal process that I'm utterly unqualified to offer you any insight).

You will need repetition, in all likelihood. Lots of it. You want to get to the point where the techniques themselves are either memorized by rote, or they're practically muscle memory to the point where the next step seems intuitively obvious once you take the one prior.

[Vector]

.

[da_nang]

Step one: Replace all roots with fractional powers. Square root, for instance, is "to the 1/2th power". Now you can use all the same rules you normally do for working with exponents. Why they even teach radical notation I don't know (maybe somebody who knows things can explain >__________>).

That's a good point. I'm pretty sure that there's no good reason for it, other than that children have a horrible time learning fractions and they want to introduce roots to entities that are assumed to be confused every time a fraction appears.

There may be some notational differences when dealing with complex numbers, methinks. Like one denotes the principal value and another the general value. Alternatively, the radical notation is more in line with "this is the solution to this polynomial". Because historical reasons.

But generally they're interchangeable, unless there is an obscure/rarely encountered detail I'm forgetting.

EDIT: Looking through some formula books, for a complex number z and positive integer n, it is defined that z^1/n = |z|^1/n e^arg(z)/n whereas ⁿ√(z) = ⁿ√(|z|) e^{arg(z)/n + 2πk/n} , k = 0, 1, 2, ... , n-1

So in this case, the radical notation is defined as the general value, whereas the exponential form is the principal. Although I suppose in the latter case, the radical of the absolute value is implied to be the principal value.

[frostshotgg]

I think the radical notation thing is just another feature of the way math is taught: based around simplifying/avoiding potential confusing things as much as possible and writing off the rest as "can't do that" or similar. At the point where radical notation is taught, students hate fractions for some reason and the concept of inverse operations isn't really explored like it needs to be for roots = fractional powers to be a thing that computes.

Year n-1: See this thing? You can't do this.
Year n: We lied, you can do that thing. Here's the long hard obnoxious way to do it now that you can understand
Year n+1: Now that you know how to do thing, here's the shortcut and a calculator that can do it for you. Here's a new thing you can do.

[Bauglir]

Ah, so there is some sort of useful distinction that radical notation's a shorthand for. Good to know - the "children are bad at fractions" thing also is possible. I don't remember well enough to recall how they were, although I do seem to recall that once we got around to algebra they clicked way better when I realized they're expressions more than anything else.

Also for anyone looking for study materials, I'm currently burning through lecture videos in the Calculus Revisited series here as a refresher because it's been like 2 years since I did math and I've got some wacky ideas that might require solving DEs and stuff (and, moreover, I've forgotten a lot of the foundational stuff, and I'm curious about it). It's pretty old, but I like the presentation for some reason. It's got some study guides and stuff as well but I can't attest to their quality since I've not taken a look.

[cerapa]

Gentlemen, I am in need of a function.



I want it to look like this, but I can't seem to figure out a way to get the result I want.

EDIT: y = ( -((x-2)^2) )/(x^2)+1 seems to be correct.

EDIT2:

Spoiler: What it was for (click to show/hide)

(they are vaguely evenly spaced becaue they are both pulling towards and away from eachother)
Function is (-square(dist-500)/square(dist)+1)*0.000001 currently.

Anyway, I was thinking about n-body simulations and wanted them to bounce into each other. So I thought I could just bake it into the gravity. If they get too close then it turns negative and pushes them away from eachother. Works pretty much perfectly. It's kinda awesome to see single high-speed particles flying around and bouncing into stuff like a billiard ball.

Actually kinda lying about the perfect thing. If some get too close by some fluke, then they bounce away from eachother at a ludicrously high speed. If I try to make a black hole, then the damn thing flies apart.