My failed attempt at deriving quadratic formula? Where did it all go wrong 🥹

[pseudo spire]

so uh

what's the general form for a quadratic eqn @dawn dawn

[dawn dawn]

hi

i got eepy

and fell asleep

[pulsar viper]

Well, if x^2 - 5x + 6 = 0, then it's not the case that x^2 = -5x - 6.

And if x^2 = -5x - 6, then it's not the case that x^2/-5 = x - 6/5.

[dawn dawn]

what?

ohhh

understood

because what is an equation?

sorry not equation

but function

you map an input to an output

My understanding was if I solve for x I could then get a solution

in reality looking at it

If I isolated the input solely on the left (x) and just put the stuff thats not x on right I would get a equation which doesnt make any sense

for example

x^2 - 6 = 0

x^2 = 6

sqrt(x^2) = sqrt(6)

x = sqrt(6)

assume x was 5

then its pure gibberish

this equation or whatever you wanna call it is garbage

I wouldnt be solving for what we intended which was, "what x's makes the y = 0"

Which in this case is none because its six below the middle...

wait

i think there are roots

for x^2-6

we start 6 down and the tails have to hit the x

[sullen hill]

Yes there are ±sqrt6 are the two roots

[sullen hill]

That's exactly how you do it except you need to find a way to isolate x in such a way that you can account for both the x² and the x

In case of something like ax² - c, it's pretty easy to see that x = ±√(c/a) for the function to take the value 0 , ie, for the roots

[mortal zenith]

Hi

[quaint zenith]

This is so easy

Just use hyperbola formula

[tardy relic]

The image isn't loading but

ax²+bx+c=0
x²+b/ax+c/a=0
x²+b/ax=-c/a
x²+b/ax+b²/(4a²)=-c/a+b²/(4a²)
= (x+b/(2a))²

x+b/(2a)=±√(-c/a+b²/(4a²))
And the rest is tidying it up

Check against this

No way it would take 3 pages

[quaint zenith]

Here -> let roots = x,y
Sum of roots -> -b/a
product of roots-> c/a
Minima of graph -> x+y/2 (equidistant)
= -b/2a
Difference of roots->
x+y= -b/a
xy= c/a
x²+y²+2xy = b²/a²
x²+y²-2xy= b²/a²-4c/a
=> b²- 4ac/ a²
Thus x-y = root (b²-4ac)/a
x-y = total distance between two roots
x-y/2 is distance of each root from center = b²-4ac/2a
Thus formula becomes->
-b/2a (mid point) +- root (b²-4ac) / 2a (distance)
-b+-(b²-4ac)/2a

There's a proof for cubic equation too! But that one is 💀

[dawn dawn]

ok

taking exam

see this

in a lil

[quaint zenith]

If you are having problem understanding I can give you a diagram too!

[placid flame]

lets not get too ahead of ourselves

[quaint zenith]

Ahahhaahhah

[errant basalt]

!tag help

[sweet karma]

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[quaint zenith]

?

[errant basalt]

Nothing

I just wanted to say

That's not allowed

[quaint zenith]

Whats not allowed

[errant basalt]

You can't just give out the solution like that

Read rules

[quaint zenith]

But others have already given the solution

And some of them used wrong methods

[errant basalt]

Wrong is wrong even if everyone is doing it, right is right even if no one is doing it

[errant basalt]

Walk through the solutions with the poster

Ask them questions

Give them hints at times

[quaint zenith]

I thought this thread was ded tbh

[errant basalt]

Mhm

[quaint zenith]

So I just wrote the solution

[dawn dawn]

yo

im here

ur interesting

wanna talk