x^2 - 5x + 6 = 0 pls explain, need help

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x = 2 or x = 3

I literally calculated that in my head. Step up, nigga.

>>8753379
((x + y)^2 - 4xy)/x - y

What about this?

>>8753387
x = 2 or x = 3
I literally calculated that in my head. Step up, nigga.

>>8753390

How did you calculate x when you were given an expression and not an equation?

>>8753374
[math] \displaystyle
\\ ax^2 + bx +c = 0 \; \; \; \; \; \; | \; \cdot 4a \\
4a^2x^2 + 4abx + 4ac = 0 \\
4a^2x^2 + 4abx = -4ac \; \; \; \; \; \; | \; +b^2 \\
4a^2x^2 + 4abx +b^2 = b^2 -4ac \\
(2ax + b)^2 = b^2 -4ac \\
2ax + b = \pm \sqrt{b^2 - 4ac} \\
2ax = -b \pm \sqrt{b^2 - 4ac} \\
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
[/math]

>>8753394
He just found its roots you stupid nigger desu senpai.

>>8753500
Is there an intuitive way to see why you apparently arbitrarily multiply by 4a and add b^2? I understand that it's done to get the factorization to pan out, but how can a non-savant see this?

>>8753520

You can not calculate roots from an expression.

>>8753374
The homework of a 12yo (solve for y)

>>8753554
No. This is how maths work. You just do stuff and hope it all adds up to what you want, no matter how arbitrary it may seem

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>>8753554
You don't need to multiply by 4a. See pic related, the superior proof of the quadratic formula.

>>8753554
here's a more intuitive way, using the method of completing the square:
[math] \begin{align*}
ax^2 + bx + c &= 0 \\
x^2 + \frac{b}{a}x &= -\frac{c}{a} \\
x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 &= -\frac{c}{a} + \left(\frac{b}{2a}\right)^2 \\
\left(x + \frac{b}{2a}\right)^2 &= \frac{b^2}{4a^2} - \frac{c}{a} \\
\left(x + \frac{b}{2a}\right)^2 &= \frac{b^2 - 4ac}{4a^2} \\
x + \frac{b}{2a} &= \frac{\pm\sqrt{b^2 - 4ac}}{2a} \\
x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\end{align*} [/math]

>>8753387
((x + y)^2 - 4xy)/(x - y)
I'm assuming you meant to write it like so^
x = y

(((x + y)^2 - 4xy)/x) - y

x = +or- y(sqrt(5) + 3)/2

>>8753390
idk how this nigga got x = 2 or x = 3; he probably shouldn't be doing this shit in his head

>>8753671>>8753374

Factorise: x^2 - 5x + 6 = 0
So: (x-2)(x-3)=0
X=2, X=3 makes zero
Thus X=2 and X=3 are the solutions

>>8753692
>x^2 - 5x + 6


I'm talking about OP's second question:>>8753387

>((x + y)^2 - 4xy)/x - y

>What about this?

>>8753671
>idk how this nigga got x = 2 or x = 3
Really?
Read the first fucking post, faggot

MUH DICK
U
H

D
I
C
K

>>8753727
I'm not talking about the question original post brainlet

>>8753390
was in response to this:>>8753387
not the original post

>>8753374
X^2-5x+6=0
X(x-5)= -6
X=-6
X-5=-6
X=-1

>>8753554

Adding the extra term or "completing the square" as it is called is actually just a generalization of a special type of trinomial called a perfect square trinomial. When I first learned about perfect square trinomials deriving the quadratic equation was obvious and I can now do it without memorizing it.

Alright.

So consider the trinomial [math]x^2 + 2x + 1[/math]

You should be familiar with it's factors, [math](x+1)(x+1) = (x+1)^2[/math]

Okay, now consider this trinomial...

[math]x^2 + 6x + 9[/math]

What are its factors?

Do you notice anything special about the relation ship between the terms of the trinomial and it's factors? The first trinomial I showed you is a trivial instance of this type of trinomial.

I'll continue...

[math]x^2 + 6x + 9 = (x + 3)(x + 3) = (x+3)^2[/math]

Hmmm.... Is there a more general way to write this? Can we rewrite the numerical coefficients in a more general form?

Well let's see... [math]6 = (2)(3)[/math] ... and [math]3^2 = 9[/math]... So I guess we could rewrite this as

[math]x^2 + (2)(3)x + (3)^2 = (x + 3)(x + 3) = (x+3)^2[/math]

And if replace all the numerical terms that are equal to each other with literal terms (as we do in algebra) we get

[math]x^2 + 2\sqrt c x + c = (x + \sqrt c)(x + \sqrt c) = (x + \sqrt c)^2[/math]

At this point you should definitely see where the extra term comes from... I'll finish explaining anyway.

So as you can see, in general, when the second term of a second degree polynomial (with a = 1) is equal to twice the square root of the third term, c, then you can always factor the expression into [math](x + \sqrt c)^2[/math]

When you derive the quadratic equation, you essentially invent a c, such that c = the square of the coefficint of b so that you can factor the left hand side of the equation into [math](x + \sqrt c)^2[/math], which in the general form of the quadratic equation would look like [math](x + \frac{b}{2a})^2[/math]

>>8754102
>coefficient of b

sorry, coefficient of [math]x[/math] in [math]ax^2 + bx + c[/math]

also all the it's should be its oops.

>>8754115
>>8754102

So, to summarize:

1. You need to be familiar with perfect square trinomials and their general form

2. You need to be comfortable replacing numbers with letters on your own

3. You need to be comfortable "inventing" terms that allow you to isolate x

>>8754102
>>8754115
>>8754119
Holy shit, I've seen a ton of explanations for solving 2nd degree trinomials over the years, but this one far surpasses them. Thanks a lot.

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>>8754007

this post gave me cancer and AIDS, thanks

>>8753374
are you trying to factor or finding the value of x?
also this is literally high-school tier question MODS!!!

>>8753374
x^2-2x-3x+6
when fully factored it becomes (x-3)(x+2)
now i'm guessing you want the zero function which would be x=3 and x=-2

easy as hell t b h you're probably hopeless if you find polynomials hard

>>8754418
(-3)(2)=6
K Y S P L E B
Y
S
P
L
E
B

>>8753554
since x^2+2x+1 is (x+1)^2
then it isn't a far-fetched idea to try to convert
the ax^2+bx into something similar by editing it a bit

>>8754418
fugg made an error, should've been (x-2)(x-3)

which means x=3, 2

Grade 9 maths lmao

>>8753374
>x^2 - 5x + 6 = 0

First you put all the numbers on one side of the equation so

x^x = 0 + 2 - 5 + 6

simplify that to x^x = 256

x triangle x is equal to 2x so 2x = 256 or x = 126. Easy.

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Mfw I see people using the ABC formula rather than the reduced, PQ formula.

>>8753705
>((x + y)^2 - 4xy)/x - y
=(x^2+y^2+2xy-4xy)/(x-y)
=(x^2+y^2-2xy)/(x-y)
=((x-y)^2)/(x-y)
=x-y
y=x

>>8753374
I used this method:
[eqn]x^2 +(a+b)x + (a\times b) = 0[/eqn]
and is easy to see that [math]a= -2[/math] and [math]b= -3[/math] so you obtain:
[eqn](x-2)(x-3)=0[/eqn]