Uh... What the fuck is x?

[Anonymous No.8697272]

Hey /sci/, I'm trying to teach myself Algebra and I shit out a practice problem that I think is too much above my level. I'd like to have some advice on this. I'm VERY confused.

Alright, so I have this problem:
>solve for X: 5x - 2x^2 + 19 = 26 + x
Alright, so I first went ahead and worked on the right side, I subtracted an x so I'd have all the x terms on the left side.
>4x - 2x^2 + 19 = 26
And then, I subtracted 19 from both sides
>4x - 2x^2 = 7
And then, I realized I'm very close to the format that factoring quadratics is in, so I decided I could try and do the equation that way. I set the equation to 0 by subtracting 7 from both sides
>4x - 2x^2 -7 = 0
Then I moved it around
>-2x^2 + 4x -7 = 0
Now it's in the proper format for quadratic factoring, but here's where my problems started.
I have never factored something like this before.

So, I need to find two factors of -7 who's sum is 4 and product is -7.
Well, -7 is a prime so I can't do that. That's where I got completely stumped. I turned to Google, and I learned I had to use the Quadratic Formula.

Okay, so I tried it, and I just got even more confused.

The formula went smoothly.
A: -2
B: 4
C: -7
Alright, so
x = -4 +- the square root of 4^2 - 4 * -2 * -7 divided by 2 * -2... Simplified it all out:
>x = -4 + or - the square root of -56 divided by -4
I can't get the square root of -56 since its negative, so I got stopped again. This time for good.

I finally turned over to Wolfram and got the answer, which looks like fucking gibberish.
>x = 1 - i sqrt(5 / 2)
>x = 1 + i sqrt(5 / 2)
How the fuck did it get this? What is i?

Can someone tell me how the hell this works? Am I in way over my head right now? I'm so confused by what I just experienced, Should I just ignore it and go back to practicing problems I can do and eventually move up to being able to do it? I'm fucking lost as hell.

[Anonymous No.8697277]

Some quadratics don't have real roots, namely those that have a negative discriminant. What wolfram gave you is the complex roots for your polynomial, where i=sqrt(-1)

[Anonymous No.8697289]

>>8697277
Oh.
Well fuck.

So what should I have done in this problem when I reached the "I can't get the square root of -56 since it's negative"
and why?

I keep looking over my work but I can't see a way out.

[Anonymous No.8697300]

>>8697289
If the delta is lower than 0 and you dont know complex numbers, just go on to the next problem

[Anonymous No.8697309]

>>8697272
>>8697289

Most algebra courses cover the complex numbers lightly, so you're learning what you're supposed to learn, remember that the sqrt of -56 can be written as sqrt(-1)*sqrt(56), we can write this as i*sqrt(56) where i is the imaginary unit, which in elementary algebra is usually defined as i^2=-1

Imaginary unit is somewhat a misnomer, they are very real with real applications, a "complex number" is a number "a+bi" where a and b are real numbers
You'll also want to note that b can be 0, or a can be 0, therefore the real numbers are a subset of the complex numbers.

[Anonymous No.8697319]

>>8697300
Fuck dude. I don't even know what a Delta is yet, According to google, it's difference or change. Is that kind of like change in y over change in x for determining the slope of a linear equation? Man I got a long way to go.

>>8697309
Well, atleast I can be reassured I'm learning what I'm supposed to learn. I think I'll go and look up more about this "imaginary number" stuff, and complex number stuff though.

All in all, frustration aside, this was a pretty fun problem to try out. So far I've been doing just intermediate stuff. It was a lot of fun and humbling to get fucked over in my steps like that. Now I have even more drive to keep on trucking forward.

[Anonymous No.8697327]

>>8697319
You may be interested to note that one of the important things you learn in elementary algebra, is that any polynomial of degree "x" (where x is the power of the highest degree term) has x solutions in the complex number system. This one had 2 solutions, like all quadratics, but they both were complex. A cubic equation would have 3, and so forth. Eventually you'll want to do problems just like the one you did, you solve it as you would any problem, but if your answer is something like "1 + or - sqrt(-25)" you pull the -1 out and say "1 + or - 5i"

[Anonymous No.8697330]

>>8697319
Oh and that other guy just used delta as a shorthand for the stuff under the radical, it's the same as the stuff you already know. Delta is also used for change in calculus

[Anonymous No.8697340]

>>8697319
>>8697330
Delta is the triangle Greek letter that I used on the photo, they teach us like this in schools

[Anonymous No.8697343]

>>8697340
My algebra courses never used delta for the discriminant, but i'm sure it's proper notation, it makes sense that OP wouldn't be familiar with it. I'm just a calc brainlet right now

[Anonymous No.8697367]

>>8697343
I'm also not familiar with it because I really just spent a month studying math(and by studying, I mean looking up broad terms in algebra, and then using khan academy to memorize how to solve equations), and then I took my GED and passed it with a 152, while I got honors on everything else.

But now, I actually kind of want to go to college and enter STEM, and that involves Calculus, so I'm taking the actual effort now to understand and do shit the right way. I've learned how to do fractions and what they represent, so I'm just moving up onto this stuff now.

So with complex numbers, I have
>3i,
so it's 3 + the square root of -1?
And 3i is just thought of a single number, made up of two parts, 3 and the imaginary number?
So it's kind of like fractions in that sense? Two parts making up a whole, only in this case the two parts 3 and i?

[Anonymous No.8697376]

>>8697367
3i equals the sqrt(-9), we can write this as 3 times i

One way I works is that it can show up as an intermediate in calculations, once you come across i^2, you can replace it with negative one, bringing you back to the real numbers

[Anonymous No.8697379]

>>8697376
3 plus the square root of -1 is 3+i

I also got a GED and had to start in algebra for college, I'm majoring in math now because I enjoyed algebra.

[Anonymous No.8697383]

>>8697367
Complex numbers consist of the real part and the imaginary part, one notation is a+b*i where a and b are real numbers and i is the imaginary unit sqrt(-1), usual visual representation is a 2d plane where X axis is the real part and Y axis is the imaginary part.

Complex numbers can be handled pretty much the same way as real numbers, they can be summed and so on, it's best to just think of them as a set of numbers greater than real numbers

[Anonymous No.8697386]

>>8697367
3i = 0 + 3*sqrt(-1)
Basically every number you ever used is a complex number. For example 16=16 +0i.

[Anonymous No.8697388]

>>8697376
>>8697379
>>8697383
alright, so suppose I have this:
>5 + 6 * sqrt(-1) + 6
So it's actually:
>5 + 6i + 6
>11 + 6i
?

[Anonymous No.8697392]

>>8697388
yes, 11+6i, if it helps, you can think of that as 11 + sqrt(-36)
When doing arithmetic with complex numbers, you can think of i as similar to x, so just as 2x + 3x = 5x, 2i + 3i = 5i, but you have to remember that if you multiply and i^2 shows up, replace it with negative 1, instead of leaving it there like x^2

2 + 3i does not equal 5i, on the other hand, just as 2 + 3x does not equal 5x

[Anonymous No.8697397]

>>8697388
to explicate on that visual representation mentioned earlier, imagine drawing a number line, now draw another number line intersecting that one (like a x and y axis) in that case, 1 to -1 is rotating 180 degrees, i would be 90 degrees, -i would be 270 degrees and so on,

This leads to a property of i which is also interesting, i^5 equals i, i^6 equals -1, and so on.

[Anonymous No.8698640]

>>8697272
Well that was interesting. Thanks for all the explanations. Still seems "irrational" logic to me though.
I focus more on proofs for basic math.
I tried to develop a better math notation system that accounted for i, but I think it's as elusive as the grandiose unification theory.
However, while I was doing this it dawned on me that x^y <> x*x...,y times. Not multiplication.
Is it possible exponents are vectors? X translated into two vectors?
Squared would be a multiple in the x axis but also in the z axis?
Is it possible we're seeing only make-believe 3d with depth?
As if we're interpreting depth exactly as we see shapes with positive or negative projection.
This is where my notation would not stabilize. Exponents would not stop exponentiating.
It was if every moment the answer would increase as if the exponential curve were ALIVE. Or I gave it the breath of life.

[Anonymous No.8699202]

> This thread isn't plagued by a bunch of faggots calling him a brainlet

Sometime you impress me /sci/
Keep working at it bro, you'll make it

[Anonymous No.8699210]

>>8699202
>tfw was just coming into the thread for the sole purpose of calling him a brainlet faggot for not knowing triple integrals before the age of 20

[Anonymous No.8699290]

>>8699210
>implying triple integrals by the age of 20 is an achievement

[Anonymous No.8699743]

>believing in (((imaginary))) numbers

oh /sci/

[Anonymous No.8699747]

>>8698640

[Anonymous No.8699830]

>>8699290
what are you Steven Hawkings?

[Anonymous No.8699864]

>>8697272

>What the fuck is x?
We have a whole board dedicated to x
>>>/x/