Quadratic Equations

[unborn scaffold]

X-a(x-b)/x-c = y --- say
rearranging i get (a+c+y)^2 -4(ac+by)
idk wht to do after this

[fickle steeple]

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[queen cobalt]

@mental gull You asked this before, right?

[unborn scaffold]

uhhh uhm.. me and him are the same..

i got locked out of that account

[unborn scaffold]

i asked then it got closed ig, no access to it

[queen cobalt]

Oh

[unborn scaffold]

💀

[queen cobalt]

Prove it via discontinuities

Rational functions are continuous and assume all real values on their domain

[unborn scaffold]

heh

he says D>=0 cuz x belongs to R

i cant understand why

[unborn scaffold]

.-.

[unborn scaffold]

pls help someone

[meager moth]

@unborn scaffold

:HelpIcon:| Help Reminder

Hello nolifeifnomath, this is a friendly reminder that your help request has been inactive for more than 24 hours. If you no longer need assistance, please consider closing the thread using the +close command. This thread will be automatically closed in 3 days if it remains inactive.

[meager moth]

@queen cobalt The user still needs help with this help request.

[covert ivy]

@unborn scaffold
https://www.mathwarehouse.com/calculus/continuity/what-are-types-of-discontinuities.php

go down to removable discontinuities

[unborn scaffold]

idk calc that much

[covert ivy]

have you done any though?

[unborn scaffold]

some

like limits and diff to an extent

(basic)

[covert ivy]

say that a = 1, b = 2 and c = 3, or the other way around, b still = 2

there would be a discontinuity at x = 2, as one cannot divide by zero

[unborn scaffold]

wht is a discontinuity

wait

ill see it rn

[covert ivy]

"discontinuity" means that the function "breaks" at a certain point

[unborn scaffold]

but idts this answers my doubt.. how can D>=0

[unborn scaffold]

quadratic never break.. i believe

[covert ivy]

either because it jumps up or because there is an undefined value

[covert ivy]

they can be

so, the point is to rearrange f(x) so that there is not such discontinuity

[unborn scaffold]

this is more confusing but i like it

[unborn scaffold]

Oh

break in the sense

it cannot input any more values right

[covert ivy]

no in the sense that there is a hole

an undefined value

or, in the sense that suddenly it jumps up

[unborn scaffold]

huh

this is the vertex of the parabola..

how is it undefined

[covert ivy]

anything/0 is undefined

if b = 2 and x = 2, x-b = 0

[unborn scaffold]

yea

[covert ivy]

thus anything/x-b = undefined

[unborn scaffold]

yea..

[covert ivy]

so it cannot be shown on a graph

the value doesn't exist on the graph

[unborn scaffold]

Ok... how is this related to my q?

my q is

How can D>=0

[covert ivy]

is D the discriminant?

[unborn scaffold]

yep

they have set x-a(x-c)/(x-b) = y
and then made it into a quad eqn

[covert ivy]

is the y supposed to be a y or an X?

[unborn scaffold]

f(x ) = y.

[unborn scaffold]

that big X was bym

its x

[covert ivy]

no

i mean the quadratic (a+c+y)^2-4(ac+by)

is the y there f(x)

[unborn scaffold]

ohh u rearrange

[covert ivy]

ok

[unborn scaffold]

the whole eqn

and that is the discrimantn

[covert ivy]

that's what i was gonna ask

so you jumped straight to determinant

[unborn scaffold]

my baddd

[covert ivy]

well

in that case

say that y = 0

[unborn scaffold]

ok..

x-a*x-c

x=a or x=

c

[covert ivy]

(a+c)^2 = a^2+c^2+2ca

and then 4(ac) = 4ac

if a or c = 0, (a+c)^2 > 4ac

if a and c = 1, (a+c)^2 = 4ac

[cobalt ether]

Um wouldn’t IVT solve this with case division

[covert ivy]

perhaps

never heard of it

[cobalt ether]

The function is continuous on (-∞,b) ∪ (b,∞)

It also has a degree 1 vertical asymptote at b, so we get that the left and right limits go to -∞ and ∞ respectively

[covert ivy]

yes, but the question i am answering is why the discriminant is >= 0

[cobalt ether]

It’s not a quadratic

[covert ivy]

they arranged it into one here

and jumped straight to the determinant

[unborn scaffold]

discriminant

pls

[covert ivy]

sorry

[unborn scaffold]

anyway

how is it D>=0

[covert ivy]

in french it's determinant

[unborn scaffold]

ohhh

mb then

its regional then cool

[covert ivy]

this is still true though, the function is a hyperbola

[cobalt ether]

Yes

[covert ivy]

we can see which is bigger if we divide (a+c)^2 by 4ac, limit of either a or c pointing to infinity

[cobalt ether]

Take b to infinity

[covert ivy]

if it is bigger than one, then (a+c)^2 is bigger than 4ac

[unborn scaffold]

dude

he says

x belongs to R

so

D>=0

im very confused

[covert ivy]

b isn't in the determinant when y = 0

[cobalt ether]

Ahhh alr

[covert ivy]

OH WAIT

makes sense

so

the determinant must be positive for its square root to produce a real number

if it is negative, the sqrt(D) will produce a complex number

[unborn scaffold]

heh

[unborn scaffold]

sq root..?

[covert ivy]

yes

when you solve a quadratic equation ax^2 + bx + c = 0, to find the two values of the variable you use:

$x_1 = \frac{-b + \sqrt{D}}{2a} and x_2 = \frac{-b - \sqrt{D}}{2a}$

[fast salmon]

Kara Ben Nemsi

[covert ivy]

finally

so

D has to be positive, also known as >=0, for x to belong to R

assuming a, b and c are real numbers, the only thing that determines whether x is real or not is whether D is >= 0 or not

that's why you were told that if x belong to R, which is a given in the exercise, D >= 0

$\sqrt{-D} = i\sqrt{D}$

[unborn scaffold]

OH

OH

[fast salmon]

Kara Ben Nemsi

[unborn scaffold]

x must real values

damn

ok got it damn

so big brain

[cobalt ether]

(x-a)(x-c) = y(x-b)
x^2 - (a+c)x + ac = xy - by
x^2 - (a+c+y)x + ac+by = 0
x = (a+c+y + √((a+c+y)^2 - 4ac - 4by))/2

Well that proves it, as the thing inside the square root is always positive by AM-GM

Hence for any y you can find 2 x that give you that y

Lmao

[covert ivy]

it's /2a, what if a = 0

[unborn scaffold]

u said was right tho

u big brain as helll

[covert ivy]

nvm, it's still defined

[covert ivy]

it just would be equal to zero

whoops

[unborn scaffold]

which

i cant understand

[cobalt ether]

Look at the quadratic again

[covert ivy]

exactly, it would still be defined, that's what made me realize what i said didn't matter

[cobalt ether]

I did an algebra mistake

[cobalt ether]

I’m getting the result that if b is large enough then there’s no solutions

What

[covert ivy]

if a = 0, a in the discriminant is just 1, always

as x^2 has a coefficient of one in the equation
x^2 - (a+c+y)x + ac+by = 0

[unborn scaffold]

i cant understand

[covert ivy]

look at what i wrote

[cobalt ether]

Yes, but with large b the discriminant becomes negative

[covert ivy]

we are solving for the quadratic x^2 - (a+c+y)x + ac+by = 0, where x^2 has a coefficient of 1, so a in the discriminant = 1

[covert ivy]

true

[cobalt ether]

$b\leq \frac{(a+c+y)^2 - 4ac}{4y}$

[covert ivy]

that poses a problem

[fast salmon]

Ravi ≥ ^• ∧ •^ ≤

[unborn scaffold]

what you told was right, no? cuz i feel it is

wait i have doubt then

[unborn scaffold]

after u get this

is it <=0 or >=0

[cobalt ether]

That’s the condition on b for the discriminant to be positive

[unborn scaffold]

he says its D<=0

💀

the soln i ean

[covert ivy]

in $x^2 - (a+c+y)x + ac+by = 0$ we have $\alpha \cdot x^2 + \beta \cdot x + \gamma$ as the template for a quadratic equation

[fast salmon]

Kara Ben Nemsi

[unborn scaffold]

yea

ok wait lemme figure this out, y is real for sure.. and D>=0

so parabola goes up

which means

y is +ve

or negative too..

cuz it intesects x axis

[covert ivy]

where $\alpha$ is the coefficient for $x^2$; in this case 1, $\beta$ is the coefficient for $x$; in this case $(a+c+y)$ and $\gamma$ is equal to $ac + by$

[fast salmon]

Kara Ben Nemsi

[covert ivy]

then, when we do $\frac{-b + \sqrt{D}}{2\alpha}$; $\alpha = 1$; so $2\alpha = 2$

[fast salmon]

Kara Ben Nemsi

[covert ivy]

alpha is never equal to 0, so this does always work, unless, once again, b is really large

[unborn scaffold]

yea

[covert ivy]

also @cobalt ether what might help is that a, b and c are in either descending or ascending order

[cobalt ether]

solve $0 \leq a^2 + c^2 + y^2 - 2ac + 2y(a+c-2b)$

[fast salmon]

Ravi ≥ ^• ∧ •^ ≤

[cobalt ether]

$y = (2b-a-c) \pm \sqrt{(a+c-2b)^2 - (a-c)^2}$

[cobalt ether]

YES

[fast salmon]

Ravi ≥ ^• ∧ •^ ≤

[unborn scaffold]

my mind

is so

confused rn

how is D<=0

[unborn scaffold]

u get another eqn

like y^2 +2(a+c-2b)y+(a-c)^2>=0
and for it go up coeff of y^2 must be positive so yea, ok and since it can give value 0 too we say its D<=0

?

[covert ivy]

c or a is always bigger than b. as shown before, (a+c)^2 is always bigger than or equal to 4ac.

[covert ivy]

nonono.

the discriminant is assumed to be always positive, that is what im trying to prove right now

[covert ivy]

this is what i said afterwards

[covert ivy]

what i was doing there is just explaining why, in these equations, the last one is divided by two only, because alpha is equal to 1, and only to 1, so the last equation always holds, as alpha cannot equal zero

$D = (a+c+y)^2 - 4(ac+by)$

[fast salmon]

Kara Ben Nemsi

[unborn scaffold]

i have

been

successfully confused

by myself

[covert ivy]

ok quick recap

[covert ivy]

D >= 0 because x belongs to the reals, so its values must be real. a, b and c are all real numbers, so the only possible non-real number would be square root of D, and square root of D is only real if D is positive, aka >=0

that is the answer to one of your questions

[unborn scaffold]

yea yea

i understand

that

but

then u get another eqn

when u do D>=0

[covert ivy]

you ignore that, don't do it

[unborn scaffold]

see from here

u have to do it

[covert ivy]

i don't think it's necessary

D is assumed to be >= 0 as x belongs to the real numbers

if D < 0; then x belongs to the complex/imaginary numbers

[unborn scaffold]

not assumed lol

we have proved its >=0

i mean u have proved

[covert ivy]

we use D only to find the possible values for x in this equation

that's it

[unborn scaffold]

but

see my message after that

[covert ivy]

this one?

[unborn scaffold]

no

[unborn scaffold]

. D>=0 is this equation

[covert ivy]

you don't need to develop the equation though

we are not trying to find specific values, we are trying to prove it can assume any value

[covert ivy]

the equation we get is this one

that is always bigger than zero since x belongs to R

and boom

that's all we need

we then plug in D into this equation:

$\frac{-(a + c + y) \pm \sqrt{[(a+c+y)^2 - 4(ac+by)]}}{2 \cdot 1}$

[fast salmon]

Kara Ben Nemsi

[covert ivy]

these are the possible values for x

that are supposedly always real

we are trying to prove that this equation always produces a real number

that is all

i was gonna work on math

but not this kind

lol

i got heavily invested

anyway, now to prove that D is always positive ignoring the fact that x belongs to R

using proof by AM-GM, as Ravi suggested:

[covert ivy]

$(a+c+y)^2 = (a+c)^2 + y^2 + 2(a+c)y$

$(a-c)^2 = a^2+c^2-2ac$

$(a-c)^2 \geq 0$

[fast salmon]

Kara Ben Nemsi

[covert ivy]

these are the assumptions

from this we can deduce that $a^2+c^2-2ac \geq 0$

consequently $a^2+c^2+2ac-4ac\geq0$

thus $a^2+c^2+2ac\geq4ac$

finally we have $(a+c)^2\geq4ac$

[fast salmon]

Kara Ben Nemsi

[covert ivy]

now we need to show that $y^2 + 2(a+c)y \geq 4by$

[fast salmon]

Kara Ben Nemsi

[covert ivy]

which can be simplified to $y + 2(a+c) \geq 4b$

[fast salmon]

Kara Ben Nemsi

[covert ivy]

@cobalt ether one question: we are trying to prove that f(x) can assume any real value, so we are trying to prove that y can assume any real value. how does this proof with which we have come up relate to that, if it does?

what we have proved so far, or shown, is the values x can take up

[covert ivy]

oh wait here is the relation

in the sense that if one x makes the equation undefined, there is always another x that will give a value, right?

which makes sense, y point to + or - infinity as x points to two

also i asked my teacher if the fact that x belongs to the real numbers proves that D is always positive, he said it's a valid proof

[covert ivy]

and this equation is defined for any real numbers a, b, c and y

so y can take any value, and there will be at least one x that exists for that value

this should normally be valid. if yes, wow. what a ride

[cobalt ether]

This isn’t strong enough

The fact that a,b,c are ascending/descending isn’t enough with that condition to prove the statement

IVT is the way I’d approach this

First of all, the equation remains the same if a and c are switched, so WLOG a < c

After that from a<b<c you get that the range of the function on (-∞,b) is a set containing (-∞,0] and the function on the interval (b,∞) has a range that has [0,∞) as a subset

QED

[covert ivy]

welp, there we go

i never imagined doing math with other people would be this fun

[covert ivy]

i just realized. they are in ascending or descending order of magnitude

[covert ivy]

does this give way to an easier/different proof?

because multiplying a number by 2 does not increase its order of magnitude, so 4by is always smaller than or equal to 2cy. If y is negative, 4by would be bigger than 2(a+c)y, so we can add that to both sides, and 4by - 2(a+c)y would remain negative, while y would remain positive, thus this equation still holds true.

so, if this proof is correct/works/is strong enough, we have proved that [(a+c+y)^2 - 4(ac+by)] is always bigger than or equal to 0

[unborn scaffold]

yea

[unborn scaffold]

yea

wait

ill send the ans? cuz im confused, i understood half part and the rest of the half part i kinda understood but its confusing

why is D<=0

[unborn scaffold]

this is so fucking confusing

how can D<=0 cuz fucking y must have real values

[unborn scaffold]

help pls

[covert ivy]

im thinking yeah

perhaps they rearranged it

[unborn scaffold]

no..

its right it seems

a frnd of mine says it has no soln

@queen cobalt sorry for ping u gotta help me

its urgent

[unborn scaffold]

@covert ivy i think wtv u told is wrong or i have confused myself

[ancient hill]

hmm

$(x-a)(x-c)=k(x-b)$

[fast salmon]

CEO of Funny

[ancient hill]

turn this into a quadratic equation of the general form @unborn scaffold

then show that k can assume any value

[fast salmon]

CEO of Funny

[ancient hill]

||you want to show that k can achieve any real value in the above equation, for real x. this is equivalent to saying that for any k, x will be real. emphasis on "x will be real"||

@unborn scaffold

[unborn scaffold]

hello

[unborn scaffold]

this all i did bro

how is D>=0

thats my doubt

[ancient hill]

oh lol

$D=(a+c+k)^2-4ac-4kb$

[fast salmon]

CEO of Funny

[ancient hill]

just expand it

and u are done

[unborn scaffold]

how is it >=0

[ancient hill]

expand it

its a perfect square

[unborn scaffold]

oh

💀 but book says

cuz all x belongs to R

so D>=0

[ancient hill]

🗣️

[unborn scaffold]

heh

[ancient hill]

But that doesnt show anything

[unborn scaffold]

ikr

same reason i said

no wait prob u getting confused

[unborn scaffold]

@ancient hill see this

[ancient hill]

,rotate

[fast salmon]

Couldn't find an attached image in the last 10 messages.

Feb 22, 2025, 7:00 AM

[ancient hill]

,rotate

[ancient hill]

cuz

$ay^2+by+c\ge 0$ implies that $D\le 0$

[fast salmon]

CEO of Funny

[ancient hill]

Cuz for the quadratic to be above the x-axis always, u need the discriminant to be negative

So that it doesnt have a root and go under

@unborn scaffold

[unborn scaffold]

Oi

i asked why >=0

[ancient hill]

they specified

cuz x real

@unborn scaffold

[unborn scaffold]

so wahttt

even in graph of a<0 D<0 (a is leading coefficient) x is all real

@ancient hill

[ancient hill]

$x^2-(a+c+y)x+ac+by=0$

[fast salmon]

miguel_125

[ancient hill]

x is root of this right?

and x is real right??

[unborn scaffold]

yep

oh

yea

[ancient hill]

sooo

[unborn scaffold]

oH

Damn

wait

wait

@ancient hill

-x^2 +ax+ b a and b are fixed say D<0, then x is also root of this

then it means

[ancient hill]

is x real?

[unborn scaffold]

obv na, x must be real cuz it shd take values

of real values..

[ancient hill]

?????????????

is the root, x, real?

[unborn scaffold]

im also confused like u

[unborn scaffold]

D<0 so no

[ancient hill]

mf im not confused

[ancient hill]

yeah exactly

now here, x, the root

is real

so D>0

[unborn scaffold]

but same case here

[ancient hill]

how so

[unborn scaffold]

ok yea true

[ancient hill]

we already know x is real na

cuz of the question

[unborn scaffold]

how is x the root then like i get it x is root

[ancient hill]

Bruh

Say x_1 is the root

[unborn scaffold]

then D<0 case which i said, x is also root there

[unborn scaffold]

its real in the qs case

[ancient hill]

Yea

[unborn scaffold]

ahhhh latex user

[ancient hill]

Yea but its not real in that case

[unborn scaffold]

yea

but

it shd input real values of x

for it form a parabola

[ancient hill]

thats the function bro

not the equation

thats f(x)=stuff not 0=stuff

[unborn scaffold]

yk what, i have struggling and breaking my mind on this q for 4 days prob thats the reason im doing shit, ill do chem and ttyl @8pm if that sounds good..

so sorry bro im very confused cuz of this q

[ancient hill]

yea sure

lol dont be sorry bro

[covert ivy]

pls dont close this thread yet

id like to go over the solutions cause its a very interesting problem that i want to write down

[unborn scaffold]

I wont lol and if i get the soln ill make sure i send it to ya xd

[unborn scaffold]

u preparing for IMO?

[ancient hill]

hmm

@unborn scaffold

imma explain the solution step by step

Now

u can read it later or whenevre u wantm

[unborn scaffold]

Wait

oh

im contemplating it lol

kk u do that

i solved it using inequalities method too ig (wavy curve)

or wait lemme be present, u explain fully then i ask the soln

[ancient hill]

Basically this is the soln.. so $x^2-(a+c+y)x+ac+by=0$ is quite clear.. now we know that $x$ is real, and since $x$ acts as a root of this quadratic, the discriminant of it should be $\ge 0$, so $(a+c+y)^2-4(ac+by)\ge 0$

[fast salmon]

wolfqz

[unborn scaffold]

dude leave this q for now..

if i have an equation x^2 + ax + b. (a and b are fixed numbers)
x is real no matter what cuz u have to input real values of x.
D<0
it doesnt intersect x axis

so any equation you take the values of x are supposed to be real

[ancient hill]

x is real no matter what cuz u have to input real values of x.

This is wrong

[unborn scaffold]

why

[ancient hill]

Why dont u understand the difference between a function and an equation

[unborn scaffold]

bcs i wasnt taught functions

💀

[ancient hill]

Oh damn

Thata crazy

[unborn scaffold]

i agree

[ancient hill]

Anyways thats no how it works

[ancient hill]

CUZ here

Its equal to 0

When ur saying "we put values of x"

its equal to y

not 0

where y is variable

[unborn scaffold]

yea

yea

hmm ok yes mb then, then still

[ancient hill]

This is an equation

U put values in a function

[unborn scaffold]

all eqns are functions na

[ancient hill]

$ax^2+bx+c=0$ is not a parabola if you graph it

[fast salmon]

wolfqz

[ancient hill]

No

[unborn scaffold]

th

why

[ancient hill]

It's either two straight lines, one straight line or nothing at all

Depending on a,b,c

[unborn scaffold]

5x^2 + 3x + 9

[ancient hill]

From the very basics of graphs.. this is like 8th grade coordinate geo

[unborn scaffold]

shd give me a parabola..

[ancient hill]

Yes

But 5x^2+3x+9=0 should not

[unborn scaffold]

oh

shit

=0

mb

fuck

yea obv it wont give you a parabola

[ancient hill]

Yess

It just highlights the roots of that parabola y=5x^2+3x+9

[unborn scaffold]

yea yea

[ancient hill]

so here if you have x^2+ax+b=0, then its not a parabola

its just the two (or less) roots of y=x^2+ax+b

So what this means

Is that if in x^2+ax+b=0, x is real

Then that means the root is real no?

@unborn scaffold

[unborn scaffold]

yes

[ancient hill]

Same thing here

D>=0 is self explanatory given the root is real

[unborn scaffold]

i get ur explanation

but

wait

[ancient hill]

Yea

[unborn scaffold]

ill send one thing tho

[ancient hill]

?

[ancient hill]

Sure sure

[unborn scaffold]

in each of these it says x belongs to r

[ancient hill]

These are not the equations

They are the functions

[unborn scaffold]

ok so?

[ancient hill]

When u say f(x)=x^2+ax+b

U can take any value of x

And f(x) will have a value corresponding to that

When u say 0=x^2+ax+b

U can only take x as some specific values

[unborn scaffold]

2 or 1 values exist

based on a and b

[ancient hill]

For which that relation works

[ancient hill]

Yup

[unborn scaffold]

but

x belongs to R..

[ancient hill]

?

[unborn scaffold]

so confusing bruh

[ancient hill]

No its not

ur confusing urself

[unborn scaffold]

hmm

[ancient hill]

Here

We have f(x)=x^2+ax+b

so when we say f(1) we mean 1+a+b

[unborn scaffold]

yep

[ancient hill]

Now

when 0=x^2+ax+b

We cant put the values of x by our choice

Like we cant put x=1 just bcz we want to

Also its not even guaranteed that x is real now

[unborn scaffold]

yea

see

[ancient hill]

?

But in our question

Its given in the start that x is real

So we use that fact to say that D>=0

[unborn scaffold]

i need some time bro so sorry, i just need to understand this a little better, i have understood but im alos a little confused

[ancient hill]

dont be sorry lmao

[unborn scaffold]

nah u giving ur time to explain

[ancient hill]

lmao dw

im free eihterway

[unborn scaffold]

i might reply only tomorrow cuz i have to do physics too 💀

u goated asff

[ancient hill]

no problemm

[unborn scaffold]

wait

can we just inequalities for this?

----a----b----c------

•       -          +           -
  

this shows u can accept all real values and get real values irrespective of the values of a b c and whether its descending or ascending

btw thats wavy curve method

[ancient hill]

it really doesnt show it

[unborn scaffold]

ah

well i gotta understand the method u sayin no matter what

kk i gotta do phy thx for helping tho, kara said the same but im a little confused still, ig got most of it cleared thx

ill come tmr and see it again

[covert ivy]

no, i just like math

[unborn scaffold]

nice

[unborn scaffold]

@ancient hill I get that since x is real, the roots are real. However that is also the range of x right..
and in this photo you can see for example the first graph.

we will take its eqn to be x^2 + ax + b.

D <0 but however he has also said x belongs to R.

@covert ivy if possible you also help

[ancient hill]

Bro

Bro

Bro

Bro

Bro

Please learn functions

Just the basic meaning

Not the entire thing

U dont know the difference between a function and an equation

Given an equation u cant just put whatever value of x, in a function you have the liberty to do that

That's why, in a function, you can put whatever real value of x you want

BUT IT WONT BE A ZERO

[unborn scaffold]

oh an eqn is technically a function where its =0?

[ancient hill]

In the equation x^2+ax+b=0

[unborn scaffold]

bruh

[ancient hill]

x must be a zero

[unborn scaffold]

i didnt know that lol mb

ok now..

(x-a)(x-c)/(x-b) is a function

and x belongs to R

according to Question

[ancient hill]

Great

$x^2-(a+c+y)x+ac+by=0$

[fast salmon]

wolfqz

[unborn scaffold]

now i equate it to 0 with that y stuff etc etc

[ancient hill]

u equate it to y

[unborn scaffold]

yea

[unborn scaffold]

mb

[ancient hill]

yeah now x is real

[unborn scaffold]

sorry

[ancient hill]

and x is a zero of that

so the zero of that equation is real

!!!

[unborn scaffold]

oh wait fuck ig i get it

holy shit

5 days

for this shit.

waw

[ancient hill]

🔥

[unborn scaffold]

its a little confusing ngl but ig i got it

wait can u give me 2-3 min to process

[ancient hill]

Yea ofc

[unborn scaffold]

we have turned a function into an equation here right

ok yea damn

i got it

waw

[ancient hill]

Hmm y is the function here

[unborn scaffold]

yea

ik

[ancient hill]

We turned it into a quadratic equation of x

[unborn scaffold]

like

[unborn scaffold]

yea yea

[ancient hill]

Yea

[unborn scaffold]

@ancient hill explain why D<=0 comes then 💀

[ancient hill]

Thats in the other equation

[unborn scaffold]

oh wait fuck thats

easy

nvm

[unborn scaffold]

yea yea

[ancient hill]

$y^2+2(a+c-2b)y+(a-c)^2\ge 0$

[unborn scaffold]

got it

[fast salmon]

wolfqz

[ancient hill]

Yea

[unborn scaffold]

cuz graph up

D<0

[ancient hill]

Ye ye

[unborn scaffold]

holy fuck bro

@ancient hill 🙏 bro had the patience and time to teach me, truly goated asf

@covert ivy thank you tho, i wont be closing so that u can note it down, however u started it off well lol, u were right,

[ancient hill]

🙏 ty ty

yw

[unborn scaffold]

i clear this one thing tho, equation is a function which has been equated to zero right?

@ancient hill

[ancient hill]

hm not necessarily

equated to anything basically

[unborn scaffold]

ye but u take that thing to the other side and its =0 now

[ancient hill]

Yeah sure

[unborn scaffold]

wait every function and equation are the same-

@ancient hill

[ancient hill]

Wym

If theres an equation smth smth =0

then u can treat that smth smth as a function

And find its zeroes

Thats equivalent

🔥

Obviously

[unborn scaffold]

no like 3x^2 =9 is a function
3x^2 -9 =0 is an equation @ancient hill

[ancient hill]

Nah

3x^2=9 is an equation too

3x^2-9 is a function

3x^2-9=0 is an equation

[unborn scaffold]

bro why am i dumb or smth

[unborn scaffold]

oh

yeaaa

thx

[ancient hill]

yw yw

[unborn scaffold]

i hope ill understand

ig i did but

takes time ig

[ancient hill]

hmmm yeah

[unborn scaffold]

thanks

[meager moth]

@unborn scaffold

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[unborn scaffold]

@covert ivy ping me when u done.

[unborn scaffold]

here he hasnt said x is real tho

[ancient hill]

hmm you'd assume it generally

[unborn scaffold]

ah

ok then

[covert ivy]

one question: if this equation is more than zero, then the discriminant must be less than zero, which is always true as shown by the following steps of the solution, correct?
we are pretty much saying this:

• Since x is real, the square root of the discriminant must be real, a a, b and c already are real.
• Consequently, the discriminant must be positive, since x is real.
• We can prove this assumption by making the discriminant itself into another quadratic equation, that is this time not equal to zero but bigger than or equal to zero
• Now, I don't understand why we automatically assume that equation's discriminant to be negative, and then try to prove it. Is it a prerequisite for that equation to be true? Why and how?

[ancient hill]

$\ge 0$ means the quadratic never goes below the $x$-axis..

[fast salmon]

LordDarkpinger

[covert ivy]

yes

[ancient hill]

if its equal to $0$, it must touch the $x$-axis and return, giving $D=0$

[fast salmon]

LordDarkpinger

[ancient hill]

if its greater than $0$, it never touches the $x$-axis, so no real roots, giving $D<0$

[fast salmon]

LordDarkpinger

[covert ivy]

thanks a lot yes

it came to me after the first part, ofc it will be complex if no real value satisfies it

k when im back home ill write down the solution

[covert ivy]

k im done writing it, you may close the post @unborn scaffold

[unborn scaffold]

kk

+close

[meager moth]

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