help-23

[feral ravine]

Wait when u square or sqrt an equation do u have to check for extraneous

Also why does vertex form exist

And why is sqrt 16 only 4 and not -4 from the get go

[fierce nimbus]

by having your quadratic equation in vertex form, you can easily find its vertex and graph it accurately

[feral ravine]

but why does it exist

like how does it show the vertex

I know what it shows/represents

but why does it show the vertex in the first place

[fierce nimbus]

from that u can find min/max of the function and a nice real life example would be finding the maximum height of a parabola when throwing a ball or something

[feral ravine]

yea but like

why does it exist in the first place

like why is vertex form the y=a(x-h)^2+k

why isn't it somethhing else

and why is h,k the vertex

im asking the WHY not the WHAT

@fierce nimbus ?

[fierce nimbus]

you have great questions but my knowledge is limited

maybe you can look at the derivation of it from the quadratic equation if that would help, but i think you want a more intuitive answer

[feral ravine]

@here is someone else able to answer

[feral ravine]

^^^

[fierce nimbus]

man i wish i knew latex

i typed it out but i think i’ll just write it down and send it so it’s better

it looks ugly on text

this is just solving algebraically but essentially vertex form stems from the quadratic equations components itself

i’m pretty sure u knew this but this is just the deriving of it

i suggest maybe watching a visual of what vertex form really stems from cause i think visualization helps way more to understand than this solving yappa

oh wait there’s meant to be a comma here mb bro

[safe radish]

@feral ravine Has your question been resolved?

[feral ravine]

@here please help me answer these questions about quadratics and square roots

how do you diffeerntiate when a square root is a function or not

cuz like sometimes u do +- and sometimes u don;t and it is confusin

and also pls answer the quesitons that are starred in the paper on the left

[polar lynx]

Don't ping random users.

[feral ravine]

i thought u were a moderator sorry

[polar lynx]

I am

[polar lynx]

What moderation issue are you pinging me for?

[feral ravine]

nvm I misunderstood your role I just need help on answering some math questions

sorry!

[feral ravine]

^^^

[polar lynx]

No worries. You should read the rules though.

We have a helper role you can ping every 15 min or so.

People with the helper role will see it. That's the whole point of the role.

Can't guarantee anybody will see or be able to help, but at least you'll be pinging the right people that way.

[dusk gyro]

what are the qeustions you have? i can't read it

[safe radish]

@feral ravine Has your question been resolved?

[safe radish]

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[visual swift]

Is the answer 4 or undefined?

[tardy mango]

4. If $f(2)$ was undefined, then $f$ wouldn’t be continuous at $x=2$.

[flat frigate]

Civil Service Pigeon

[warm warren]

if its undefined, then it wont be continuous. Maybe simplifying is the correct method.

[tardy mango]

Note they’re essentially asking you to fill in the blank for
$$f(x)=\begin{cases} \frac{x^2-4}{x-2}, & x \neq 2 \ \boxed{?}, & x =2 \end{cases}$$

[flat frigate]

Civil Service Pigeon

[tardy mango]

Which you correctly did with 4

Idk if the second part not being explicitly written is what confused you

[finite igloo]

https://tenor.com/view/shocked-cat-slowed-bird-gif-2022555196927695723

[visual swift]

There are two interpretations that I’m debating between:
x not equal to 2 is connected to the f(x)=…
Or it’s separate, so 2 isn’t in the domain

[finite igloo]

.

[vagrant ice]

terribly worded problem

[finite igloo]

2 isnt in the domain initially

but like others has said

they want u to extend f so that f is continuous

[vagrant ice]

should be worded like this

[light shoal]

"f is continuous on all real numbers", so 2 is definitely in the domain of f, it's just not in the domain of the formula they gave

[vagrant ice]

for which value of b is $f(x)$ continuous, if $f(2) = b$?

[flat frigate]

south

[finite igloo]

i believe the question's still poorly worded nonetheless ;-;

[light shoal]

yea i agree that it could be worded better, but it's not outright wrong at least 🤣

[visual swift]

I answered 4 with the same explanation you guys were saying, but my teacher said otherwise, and that x not equal to 2 isn’t connected to f(x)=…

💀 what can I do to convince them

[finite igloo]

BRUH

ask ur teacher

if f(2) is undefined

how tf f is continuous for all real

🥀 🥀

ts pmo

[visual swift]

What do I do to convince them? 💀

They’re saying that x neq 2 is an addition on the statement that f is continuous for all real numbers

[safe radish]

@visual swift Has your question been resolved?

[warm warren]

what was the answer your teacher provided?

[visual swift]

Undefined

So answer choice d)

[visual swift]

^ basically like the x not equal to 2 served as {x| x neq 2}

That was the reasoning

[warm warren]

how is it continuous over R if its undefined at 2?

[safe radish]

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[visual swift]

.reopen

[safe radish]

✅ Original question: #help-23 message

[visual swift]

They’re saying that x not equal to 2 modified the f is continuous statement

[obsidian oracle]

"f is continuous at all real numbers" is the same thing as "f is defined and continuous on R"

The fact that you're given a formula for f(x) for x ≠ 2 doesn't remove the fact that f is also defined at 2, which is what the first statement says

It's just that f(2) is not given, it's up to you to determine

[obsidian oracle]

Explain to your teacher that if the first statement says f is defined on R, then f is defined on R no matter what the next statements are
The next statement only states what f(x) is when x ≠ 2, and doesn't say f(2) doesn't exist

Imagine I have a function f that determines the eye color of Alice and Ben
And I say "f(x) = blue if x ≠ Alice"
Does that make Alice's eye color undefined?

Weird example but maybe it shows better the absurdity of saying the answer is d

[visual swift]

Thank you so much! I’ll make up what I say according to this.

.close

[safe radish]

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[kindred venture]

So i have 3 chart i need to understand

[kindred venture]

yuh its math related <@&286206848099549185>

[quasi timber]

Dude

[kindred venture]

yes 😩

[ionic tide]

Yes mamm

[quasi timber]

😭 what do you want help with all im look at is just a pile of graphs

[kindred venture]

i need help what are those graphs?

😭

[quasi timber]

idk they don’t look math related at all.

[kindred venture]

what do they represent + explain

[quasi timber]

this is physics i think.

[kindred venture]

co ordinate geometry?

[quasi timber]

oh good fionna please help me

[lost jewel]

The last one is Chemistry

Hell nah

[quasi bison]

wtf kind of incompetent teacher gives you a bunch of graphs with zero context and demands that you explain them

[lost jewel]

the last one has that wavelength chart that I'm so scared of

[kindred venture]

my teacher

😭

[quasi timber]

get their ass off the education board

[quasi timber]

btw !15min

[kindred venture]

wha

[safe radish]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[spring yoke]

no ping helpers for 15 min

[kindred venture]

!close

💔

useless 🥀

[spring yoke]

.close

cmd

[kindred venture]

.close

[safe radish]

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[kindred venture]

.bye

[safe radish]

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[jolly kelp]

could someone plz explain to me in detail how the b with the root of 4 gets moved down to the denominator with a cubed b now inside

so going from applying th quotient property of logs to simplifying. cause i thought the answer after applying the quotient property was it, didn’t think i could simply any further

[native flume]

yh il send a ss one sec

[quasi bison]

$\frac{\sqrt[4]{b}}{b} = \frac{b^{1/4}}{b} = \frac{1}{b^{3/4}}$

[flat frigate]

Ann

[quasi bison]

nothing is really "moved" as such @jolly kelp

[jolly kelp]

this is great

okay so it’s an operation of fractions as exponents between the nominator and denominator

[quasi bison]

numerator*

[jolly kelp]

not to be rude but written like this might as well be a foreign language

[quasi bison]

you should not be looking at my message

but rather at the picture that the bot sent immediately beneath

[quasi bison]

ie right here

[jolly kelp]

that makes more sense

ya got you dude didn’t realize that’s what you were doing

and

thank you guys for the kind explanations

[safe radish]

@jolly kelp Has your question been resolved?

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[native flume]

hi could i get some inspiration for this question. ive tested a few small cases, and n = 2 and n = 3 work. ive also said that all squares can be in the form 4k^2+4K+1 or 4k^2, i thought that 4k^2+4k+1 is similar-ish to n x 2^n +1, so im trying to consider when n x 2^n can be in the form 4k^2+4k+1, (this is probably going the wrong direction) but then i said that n x 2^(n-2) = k(k+1) (not useful i think) bit stumped. are there any concepts i should be thinking abt, thanks

[tardy mango]

$n \cdot 2^n+1=k^2 \implies n \cdot 2^n=k^2-1$

[flat frigate]

Civil Service Pigeon

[tardy mango]

How can you ||rewrite k^2-1||

Further hint: ||consider the 2-adic evaluation of both sides, aka the greatest power of 2 that divides both sides||

[native flume]

ah difference of 2 squares

ty il keep trying

[tardy mango]

[safe radish]

@native flume Has your question been resolved?

[native flume]

so i (we) established that n x 2^n = (k-1)(k+1). i noticed that n can be either odd or even, and that 2^n is always even, hence, their product is even. therefore, (k-1)(k+1) are both even. considering if n is odd, in the case n = 3, we have 3 x 2^3 = (k-1)(k+1), but since 3 is odd, we must give it a factor of 2 from the 2^3, hence, we get 6 x 4. for n = 5, we have 5 x 2^5, again, we give 5 a factor of 2, so we get 10 x 16, since this isn't a difference of two, we have to give 10 another factor of 2 (this is the only thing we can give it since 16 is in the form of 2^n, which only has factors of 2). i tried this for more odd n's, and apart from 3 this seems to never happen again, but i still havent found a rigorous proof for this. i then tried doing it if n is even. for example, if n = 2, we get 2 x 2^2 = 2 x 4 which works. if n = 4, we get 4 x 2^4 = 4 x 16, only with the same logic as before, we have give a factor of 2 from 16 to 4, which gives us 8 x 8, i then tried this for a few more, and it seems like we can never have n x 2^n equal to two numbers with a difference of two, but i feel like this isn't that rigorous and perhaps there could be larger numbers of n such that n x 2^n does equal (k-1)(k+1)

i just saw the further hint acc i think i can keep going

[tardy mango]

[safe radish]

@native flume Has your question been resolved?

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[safe radish]

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[dawn token]

Hi ! I'm on an exercice about polynomial.
The exercice is divied in 2 questions. I achieve the first part of the first question :
"Determinate a polynomial of degree 2 such that P(- 1) = 1, P(0) = 1 and P(1) = 1."
And I found P(X) = 1/2 X^2 - 3/2 X - 1

But, I don't understand how to say if "Is this polynomial unique?" or not.

(Sorry if some terms aren't good ; I'm used to talk in english, but rarely about maths).

[marsh walrus]

Sorry, you have P(1) = 1 twice here.

[dawn token]

ah, sorry, one of those is P(-1) = 1

[dawn token]

it's done (I changed it).

[marsh walrus]

It depends on the class I think how you would do this

Do you have an opinion about if its unique or not?

[dawn token]

I think it's not unique, no because I feel it, but because the following question is "Determinate all the polynomial P ∈ R[X] such that P(−1) = 1, P(0) = −1 and P(1) = −1."

[marsh walrus]

It is possible that theres only 1, right?

or maybe there are none

[dawn token]

You mean, degree ? I don't know, I just copy past the question ; I have no others information than what I have already written.

[marsh walrus]

did you change all of these to 1 on purpose?

P(-1) = 1
P(0) = 1
P(1) = 1

[dawn token]

ah, no.
It's :
P(-1) = 1
P(0) = -1
P(1) = -1
(sometimes the sign disappear, I don't know why).

And I changed nothing from what is it written in the exercice.

[marsh walrus]

if thats right then this polynomial isnt correct

did you want to go through solving for the polynomial?

or we can just talk about the second part

[dawn token]

I already solve it... I think ? (for the first part of the question ; I wrote my answer above). So I don't mind about talking about the second part, if it's the question "This polynomial is unique or not."

I found that :
P(X) = 1/2 X^2 - 3/2 X - 1

[marsh walrus]

,w graph x^2/2 - 3x/2 - 1 from x=-2 to x=2

[flat frigate]

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[marsh walrus]

,w graph x^2 - x - 1 from x=-2 to x=2

[flat frigate]

Results provided by WolframAlpha

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Requested by janniku

[marsh walrus]

just something to think about

[marsh walrus]

What kind of method did you use to solve it?

I mean, sometimes, people write out a general polynomial like $ax^2 + bx+c$ or something

[flat frigate]

jan Niku

[marsh walrus]

and then you use the points to get three equations

[marsh walrus]

Yea, exactly

I mean, exactly to the method

you have three equations, and three unknowns

each time you 'use' a point, you can get rid of an unknown

so once you have three points, for three unknowns, theres no more room for many different answers, you have exactly solved each unknown

[dawn token]

yeah, I understood that, but I don't understand how to know if this polynomial is unique or not. (And whatever means "unique" for a polynomial).

[marsh walrus]

it means, if there is another polynomial that also go through those 3 points

like, say we are talking about one point, (0, 0)

which line goes through (0, 0)

well 2x does

but its not unique

any line mx, for any m, will go through (0, 0)

[dawn token]

AH ! okay !

[marsh walrus]

what if we add another point?

so now 2 points, which line goes through them, is it unique?

maybe you have explored this one more than the quadratic

but you can also just imagine, you have two points, and you connect them with a line

you don't get any choices

[dawn token]

okay.... I think I understand. There 2 points to check, and there can be only 1 answer to it, because it's a direct line.
I understand that. But I don't see the link with that and the question about polynomial.

[marsh walrus]

So, 2 points determine a unique line, right

this reflects the fact that some line is ax+b

two unknowns, a and b

so, two points, and its completely determined, you don't have any more choice

This is just like a quadratic, ax^2 + bx + c

Do you buy it? @dawn token

or still doubtful

[dawn token]

I still don't understand.

Do you have an exemple with a polynomial not unique and another unique ?

[marsh walrus]

like a quadratic?

[dawn token]

For quadratic, yeah.

[marsh walrus]

say P(-1) = 0 and P(1) = 0

they're roots, so we can construct this one pretty easily

one solution is (x-1)(x+1) = x^2 - 1

,w graph x^2 - 1

[flat frigate]

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Requested by janniku

[marsh walrus]

can you find another solution?

like I was arguing, a quadratic has three unknowns, but we're only using 2 points, so there should be more

[dawn token]

Okay, for now.

[marsh walrus]

How do you mean?

[dawn token]

I mean, I understand your explanation until now. If there are 3 points on the graphic, it means more solution ?

[marsh walrus]

yea

[marsh walrus]

like here for example

2(x-1)(x+1) still has roots at -1 and 1

so does 3(x-1)(x+1)

theres actually infinitely many quadratics that goes through those points

[dawn token]

And we know that how ?

By knowing by heart ?

[marsh walrus]

we can check

try $P(x) = a(x-1)(x+1)$

[flat frigate]

jan Niku

[marsh walrus]

we need that P(1) = 0 and P(-1) = 0

$P(1) = a(1-1)(1+1) = a \cdot 0 \cdot 2 = 0$

[flat frigate]

jan Niku

[marsh walrus]

$P(-1) = a (-1-1)(-1+1) = a \cdot (-2) \cdot 0 = 0$

[flat frigate]

jan Niku

[marsh walrus]

with me so far?

[dawn token]

Yeah, I still understand.

[marsh walrus]

we notice now, it never mattered what a was

we can make a any number we want, and these equations are still true

[dawn token]

As long as his form is a(x-1)(x+1), it doesn't matter who is 'a' ?

[marsh walrus]

yea

you ever use desmos?

[dawn token]

desmos ?

Never heard about it.

[marsh walrus]

https://www.desmos.com/calculator/iy2z9fqvap

we need P(x), the red curve, to go through the black points

on the left, try moving the slider for a

[dawn token]

It always pass by the 2 points ?

[marsh walrus]

it does yea, no matter what we make a

this is because P(x) isn't unique here

and that's because you can't uniquely solve for three unknown variables with only 2 pieces of information

[dawn token]

We prove it only by making another P(X) with a different from what we found ?

[marsh walrus]

that's one way to show it's not unique, yea

[dawn token]

So, it's not unique, only if there are 2 pieces of information to find what we search ?

[marsh walrus]

right

because some just like, generic, any quadratic is $P(x) = ax^2 + bx + c$

[flat frigate]

jan Niku

[marsh walrus]

three unknown variables a, b, c

the reasoning goes: three unknown variables, you need three pieces of information to solve them

just like with a line. Give me two points, and connect them, there's only one unique line that goes through them

[dawn token]

... It's impossible to get a not unique polynomial when his form is :

$P(x) = ax^2 + bx + c$

[flat frigate]

Jiraknight3

[dawn token]

?

[marsh walrus]

Is it possible to get a unique polynomial?

The one in your original problem is unique

[dawn token]

I mean : with this form (above), there's no way that can be no-unique ?

It will always be unique, no matter what, right ? or not

[marsh walrus]

It depends on what information you have

when we write ax^2 + bx + c, we mean, a b and c can be any number

then if we have information, we can start to determine what a b and c must be

for example, if they say P(0) = 0, we know that c = 0

so if P(0) = 0, then P(x) = ax^2 + bx

where a and b can still be anything

[dawn token]

yeah.

And if we have two others informations, like... P(2) = 4 and P(6) = 9 ?

Just for example.

[marsh walrus]

then we can totally solve it

[dawn token]

And after solving it, how will we be able to prove that it's unique, or not ?

[marsh walrus]

There are a few ways to do it, it depends on how much math you have

I can explain to you one way to reason it with just algebra

You have to know one thing first

if we add or subtract multiple quadratics, we can only get a quadratic (or less) as a result

[dawn token]

I am in the first year of a computer science degree, so I still have quite a bit of math in my program.
I would like to.

[marsh walrus]

if you have linear algebra its a little easier

[marsh walrus]

I mean, if you have two degree 2 polynomials, and you add or subtract them, the result is degree 2 or less

do you believe it?

[dawn token]

Linear system and gauss method ?

[marsh walrus]

yea

[dawn token]

We didn't do it yet.

[marsh walrus]

thats okay

[marsh walrus]

you can use this

[dawn token]

Can you illustrate it ?

With an exemple, I mean ?

[marsh walrus]

$P(x) = ax^2 + bx + c$ and $Q(x) = dx^2 + ex + f$

[flat frigate]

jan Niku

[marsh walrus]

then $P(x) + Q(x) = (a+d)x^2 + (b+e)x + (c+f)$

[flat frigate]

jan Niku

[marsh walrus]

So, we add P and Q, we cannot get a cubic, or a quartic

we can't create x^3 or x^4 or x^5 or anything higher

[dawn token]

Yeah, okay.

[marsh walrus]

if P is degree 2, and Q is degree two, then P+Q is degree 2 or less

[dawn token]

Yeah, I saw that.

[marsh walrus]

okay so now lets look at your problem

you are given 3 points

You found one solution, I think it's unique

let's say I'm wrong, and you have another solution

So we have both P and Q, they're degree 2 polynomials, and they both go through those 3 points

that means that P(1) = Q(1) and P(0) = Q(0) and P(-1) = Q(-1)

[dawn token]

We invented the other solution ?

[marsh walrus]

I'm arguing that you have the unique solution

one way to show I'm right so to do this

Say I'm wrong, and there is another solution Q.

then, we want to arrive at something that cannot be true

[dawn token]

It's a proof by absurd ?

[marsh walrus]

Yea what is it uh

reductio ad absurdum

something like that

we call it contradiction

[dawn token]

Okay.

[marsh walrus]

So we have another solution

P and Q are both solutions

P(-1) = Q(-1) and P(0) = Q(0) and P(1) = Q(1)

and possibly at other points, they are not the same

What happens if we look at P(x) - Q(x)?

we don't know exactly what it might look like

[dawn token]

There will be equal to 0 ?

[marsh walrus]

it will in some places, yeea

[marsh walrus]

like, they are equal at these points

so P(x) - Q(x) must be 0 at -1

and it has to be 0 at 0

and at 1, it also has to be 0

so we have a polynomial, P(x) - Q(x)

and its 0 at three points

do you see the problem?

[dawn token]

I follow actually and understand.

[marsh walrus]

Can you finish out the logic?

P is degree 2

and Q is degree 2

[dawn token]

Not at all...

[marsh walrus]

we can go back to this

P is degree 2, and Q is degree 2

so P - Q is degree 2, or less

[dawn token]

Yeah.

[marsh walrus]

Can we make a quadratic have three roots?

maybe you know the quadratic formula

[dawn token]

yes

[marsh walrus]

it tells us two roots

[dawn token]

I thinks ?

[marsh walrus]

nope we can't do it

its shaped like a U right

i mean, we don't have to argue shapes

the quadratic formula tells us all of the roots of a quadratic

there are at most 2

do you believe that?

[marsh walrus]

I mean, its really this again

in a different shape i guess

if you want P(x) to have roots r1 and r2 then $P(x) = a (x-r_1) (x-r_2)$

[flat frigate]

jan Niku

[marsh walrus]

if we want to add another root wed need $P(x) = a(x-r_1)(x-r_2)(x-r_3)$

[flat frigate]

jan Niku

[marsh walrus]

but now P is a cubic!

[dawn token]

r is for racin ?

[marsh walrus]

r is for root

[marsh walrus]

if this doesnt make sense dont worry too much about it

[dawn token]

yeah, ok, racine (FR) = root. Understood.

[marsh walrus]

oh, interesting

a quadratic cant have 3 roots

So that's the main point of the logic

You can maybe finish it out, if you want the practice

The last piece is that we can actually make P(x) - Q(x) be 0 at those three points

if Q is just P

then P-Q is 0 everywhere

Maybe you can see the problem this raises, if you want to argue that P is not uniquE?

I hope this is making some sense

[dawn token]

I'm really trying to understand...

But, I don't see where that lead.

[marsh walrus]

We want to show P is unique by contradiction

So assume that there exists another quadratic, Q, such that it also goes through those three points.

Since P is a quadratic, and Q is a quadratic, then P - Q is a quadratic (or less).

However, P - Q = 0 at x= -1, 0, 1.

@dawn token good to here?

[dawn token]

yes, that, I understood.

[marsh walrus]

Okay, and you believe that a quadratic cannot have three roots?

[marsh walrus]

Commonly, in the USA at least, we go through getting this formula, which tells us all (both) the solutions to P(x) = 0

[dawn token]

A quadratic cannot have 3 roots = it's a fact ! right ?

[marsh walrus]

Well we can write them both down.

[marsh walrus]

it's a fact, yea

but hopefully you're comfortable that it's true

if not, I'd encourage you to try to build a quadratic with three roots

Maybe you would try to draw one, and then pick some points on your drawing, and solve for the equation

or use desmos

[dawn token]

Someone that I met who was very good at math tell me to not understand and just learn equations or facts like that, to be good at math.
But, from that... what do we do ?

[marsh walrus]

I think you have all of the pieces of the proof

all that you need to do now is to convince yourself

Or, maybe you don't understand the structure of the proof and we can talk about it more

it's good to play with things, or try things and be confused, I think

[dawn token]

I don't understand how to apply that for my exercice...

[marsh walrus]

Let's assume you believe this thing about quadratics is true

that a quadratic cannot have three roots

[dawn token]

Yeah, I suppose there are a Q(X) who is equal to P(X) for X = 1, 0, -1
But, I know nothing about this Q(X) to do anything like P(X)-Q(X).
I just know P(X) that I found earlier.

[marsh walrus]

We know some things about P(x) - Q(x), right?

[dawn token]

I can do nothing with this Q(X), concretely.

[marsh walrus]

because they're equal at x=1, 0, -1

so, for example, what P(0) - Q(0)

[dawn token]

this is equal to 0, yeah.

[marsh walrus]

its not 0 everywhere

damnit

<@&268886789983436800>

[marsh walrus]

its not 0 everywhere, no

In the contradiction we're hoping that P and Q are different functions

[dawn token]

only for X = 1, 0, -1
But, how can we prove it that it's not 0 everywhere ?

[marsh walrus]

in the contradiction

we're assuming that there is some other Q that is also a quadratic that goes through those three points

[dawn token]

Yeah, P and Q are differents.

[marsh walrus]

possibly, yea

at least, they aren't necessarily the same

if P isn't unique, by assumption, then we must be able to find some other Q

[dawn token]

And how can we try to do that ?

[marsh walrus]

We are just assuming it is true

[dawn token]

But, we can do nothing with this Q, for now, to prove it that P is unique or not.

[marsh walrus]

we can

<@&268886789983436800>

All we need to know is that Q is a quadratic, and that it also goes through those 3 points

[dawn token]

Okay...

[marsh walrus]

okay so we have P and Q

they are maybe not the same

but they both go through the same points at x=-1, 0, 1

now we are interested in P(x) - Q(x)

This is also a function, and it's a polynomial, of degree 2 or less.

[marsh walrus]

You remember we talked about this result?

[dawn token]

Yeah, I understand until now...

[marsh walrus]

ah crap I got to take a phone call

[dawn token]

No prob.

[marsh walrus]

Okay, so what is P(x) - Q(x) at x=-1?

so what is P(-1) - Q(-1)

[dawn token]

0

[marsh walrus]

yea

and at x=0 and x=1 right

[dawn token]

Yeah.

[marsh walrus]

So P(x) - Q(x) might be some wacky quadratic, we don't know exactly how it looks necessarily

but we know that it has three roots, a x=-1, 0, 1

[dawn token]

And, from that, we do what ?

're you here ?

@marsh walrus

(I wait, no problem).

[marsh walrus]

@dawn token yea, sorry

[dawn token]

no prob.

[marsh walrus]

I'm an on call guy at my job and I get three phone calls a year lol

they're all coming in today

[marsh walrus]

Now we have enough information for a contradiction

We have a quadratic, P(x) - Q(x)

and it has three roots

You don't seem convinced of this though

[dawn token]

I understand everything until now, but, I've literrally 0 idea what to do with all these information, to do a contradiction.

[marsh walrus]

Well we are in a tough spot now

because we want to make this work

for P - Q to have three roots

Quadratic formula tells us, a quadratic can have at most 2 roots

[dawn token]

AH §

[marsh walrus]

we can also write out what P - Q must look like to have these roots

[dawn token]

WAIT !

[marsh walrus]

it must look like ax(x-1)(x+1)

but this isn't even a quadratic

[dawn token]

huh, if I call P - Q, let's say, "F"
F is a quadratic.
But, we know at the same time that F, as a quadratic, cannot have 3 roots.
But, F = P-Q and P-Q have three roots.
So it's impossible !

right ?

Or I still don't understand ?

[marsh walrus]

Yea, that's more or less the idea

There is still one more option

If we really, really want to make it work

what if P - Q = 0

so, the F you are talking about, if it's just 0 everywhere

then for sure F and P-Q are 0 at x=-1, 0, 1

[dawn token]

ah, yeah...

[marsh walrus]

because they're zero everywhere

Does this work?

what if P - Q = 0 everywhere

[dawn token]

that means P and Q are equal ?

[marsh walrus]

Yea

so, we didn't find a new solution, we just found P again

[dawn token]

So, from that, P is for sure unique ?

[marsh walrus]

Yea, that's our argument

if there are 2 quadratics that go through those three points

it must be that they're the same

[dawn token]

I thought I won't never be able to understand.

[marsh walrus]

once you do more linear algebra you will have easier ways to do this

I don't know that your problem is asking you to write a proof either
but since you were curious

I imagine they want you to just say something like

because there are 3 points, and 3 points uniquely determine a parabola

[dawn token]

Can I try, from what we see there, to resolve my exercice, and to ask to you if what I have write it's true ?

(To be sure I understand it).

[marsh walrus]

sure

[dawn token]

Thanks ! I will do that right now ! I will ping you when I finish.

[marsh walrus]

i may end up getting another phone call, so if I am not responding just give me a moment

sure

[dawn token]

No problem.

Some of sentences are in french ; but the essential is here and it's understandable, normally with all the numbers and letters. @marsh walrus
(the last sentence is just "This polynomial is unique").

[marsh walrus]

It's a little blurry but yea I think you have the idea

[marsh walrus]

hopefully this is overkill for the assignment haha

or, maybe they wanted you to show that you can row-reduce it to the identity

[dawn token]

As long as I manage to respond and pass my exam, that’s fine for me. haha

[marsh walrus]

Well thanks for sticking it out, sorry that was a long one haha

let me know if you had any other confusions

[dawn token]

I readed that I needed to create another channel if I had another question...
(Because I have another...)

[marsh walrus]

I would suggest to make another channel, so that you are near the top

and if I can't help you have a better shot of getting someone else

people are afraid of the older channels lol

[dawn token]

Okay. Thanks. I will do another channel, then.
Big thanks again ! You're truly a sweetheart ! ❤️ I've never thought that I will be able to understand it.

.close

[safe radish]

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[safe radish]

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[daring river]

help

[daring river]

Hello

I need help for

[keen tulip]

Hello

[keen tulip]

Thoughts? What have you tried?

[daring river]

I tried to find out where it increases and decreases and then foudn otu the local max and min

Is that right?

[keen tulip]

Yes, that's a good approach

What have you got so far?

[daring river]

Like what increases and decreases?

And whats the max and min?

• infintie, -0.618 increases
  -0.618, 0.5 decreases
  0.5, 1.618 increases
  1.618, infinite decreases

??

[regal onyx]

thats good

did u plot the points they gave u

[daring river]

ye but idk if thats right

If I send a pic can u tell me if its right

or not

[regal onyx]

ye show me what u have

[daring river]

.close

[safe radish]

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• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
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[hoary seal]

just wondering what is a_n

[hoary seal]

in this context

[regal onyx]

term number of the geo sequence

[austere goblet]

nth term

[cedar widget]

the n'th term of the sequence

[hoary seal]

what is nth term?

[regal onyx]

1st term n=1

2nd term n=2

5th term n=5

so nth term

[hoary seal]

oh its a generally defined

I thought its the end number

[austere goblet]

yes

[hoary seal]

ok make sense

[outer willow]

How can i ask for help

[hoary seal]

!help

[safe radish]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[outer willow]

!help

[safe radish]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[outer willow]

oh

[hoary seal]

.close

[safe radish]

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Available help channel!

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• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
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[jolly crescent]

can someone help me with this ratios question
find the ratio a : b : c
a : b = 2 : 5 b : c = 4 : 3
i have a test soon and i dont get this an someone help me?

[haughty glacier]

in that ratio

a would = 2

since its corresponding

its 1st in the ratio adn that number is first in that ratio

then u just do that to fdind the others

and give the ratio

wait no im wrong

sorry

[snow robin]

you want to make it so that b is both 4 and 5

since b is 5 in one of the ratios

and 4 in the other

[jolly crescent]

yeah

[snow robin]

you want to find a way to represent b as both

you can multiply both numbers of a ratio by another number without changing the ratio

as 2 : 5 is the same ratio as 4 : 10

this means you could change the ratio 4 : 3 to be 5 : something

but that would not be whole numbers

it would work

but it wouldn't be clean

to do it cleanly you want to find a way to make b into 5 and 4 at the same time

how do you think you would be able to do that?

[jolly crescent]

u times it by a

u times it

to make it the same number?\

[snow robin]

yes you want both ratios to have b as the same number

but how would you do that but still keep whole numbers

[jolly crescent]

im not usre

[snow robin]

what if you multiply the first number for b with the second?

so multiply the first ratio 2 : 5 with 4

and the second 4 : 3 with 5

[jolly crescent]

so 2:20 and 20:3

[snow robin]

no because now you've changed the ratios

2 : 20 is not the same ratio as 2 : 5

[jolly crescent]

oh

[snow robin]

you almost have it

you only multiplied one side of the ratio

instead of the entire thing

[jolly crescent]

so i multiply everything

8:20

and 20:15

?

[snow robin]

yes

[jolly crescent]

so its 815

.

8 : 20 : 15

[snow robin]

yeah

you did it

[jolly crescent]

Ohh thats the final?

tysm

[safe radish]

@jolly crescent Has your question been resolved?

[safe radish]

Channel closed

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[indigo scaffold]

31 000 = a/0-(-10) + 35 000

[indigo scaffold]

-4000 = a/10

-40 000 = a

?

.close

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• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
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• Be polite and have a nice day!

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[vast shard]

can someone help please.. idk what to do

[vast shard]

im soooooooooooooooooooooooooo bad at integration

because i mainly dk what method to choose to solve

[vestal carbon]

Do you know what partial fractions are

@vast shard

[vast shard]

yes, i tried doing that

but

since the denominator isnt factorable i wrote

Ax+B/ denominator

idk how thats gonna help me

[woven hound]

factor x^2 - 4x + 5

you don't want Ax+b just yet

[swift fjord]

you cant factor it

GPT says to complete the square and that works out cleanly lol

[vast shard]

yea i cant

[vast shard]

whaat

ive never learned that method

[swift fjord]

its algebra 1

with ax^2 + bx + c

u just write in as (a+_)^2 + _

and to do that u base it off b

a + k ^ 2 = a + 2k + k^2

we want 2k = b

so

k = b/2

idk u can ignore the work but basically u rewrite it as

(x-2)^2 + 1

[woven hound]

!noai

[safe radish]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[swift fjord]

completing the square just means to rewrite the quadratic with a squared binomial and a constant

[woven hound]

oh you're right I misread the problem mb

[swift fjord]

maybe you need ai

[swift fjord]

anyways from here its just a u sub with u = x-2

lowk this is a pretty unintuitive integral, hard to find

[grave lintel]

hi

i'm new

[twilit spindle]

#❓how-to-get-help

[swift fjord]

@twilit spindle no way u see completing the square allows you to u sub which allows you to split the integral into ln and inverse tan lmao

maybe its been too long since bc but idk seems pretty difficult to notice

[twilit spindle]

well I mean split x-1 into two fractions

1/x^2 smthn obv arctan

x/x^2~1/x smthn

obv ln

also hello again cat

[safe radish]

@vast shard Has your question been resolved?

[vast shard]

yea is there an easier approach?

[vast shard]

how did you come to this conlcusion?

[vast shard]

and this conclusion?

[vast shard]

hello hello

im surprised you remember me ;0

[swift fjord]

its not an easy conclusion

[vast shard]

does it really just have to come through practice 😔

[swift fjord]

but probably slightly easier than completing square first then u sub first then splitting

if u split first

like 1 / x^2 + 4x + 5

u can kinda guess inverse tan

or at least consider it

its not clean cuz it still becomes two terms but

idk

¯_(ツ)_/¯

[vast shard]

bruh

i thought inverse tan was

[swift fjord]

hes saying that 1 / x^2 + ... usually is inverse tan (which i dont even think is generally true)

[vast shard]

smth/ (variable)^2 + (constant)^2

[swift fjord]

and x / x^2 + ... usually ends up with ln (probably more true)

[swift fjord]

yea exactly

[vast shard]

hmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm

[swift fjord]

thats why completing the square is useful

it turns a quadratic into

smth ^ 2 + a constant

[vast shard]

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

okoay

makes sense

[twilit spindle]

,w int 1/(ax^2+bx+c)dx

[flat frigate]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by xxmrfancyu2xx

[twilit spindle]

t'is generally true

[vast shard]

integration sucks

[plucky hearth]

i remember putting this in wolfram

[twilit spindle]

course I did!

[swift fjord]

not if the denom is factorable

[twilit spindle]

well if we have a x over x^2, then that looks like 1/x

[swift fjord]

or even just 1/x^2

[twilit spindle]

no that still works irregardless

[swift fjord]

yes

but

[vast shard]

yippie!

[swift fjord]

thats not how u wanna solve those

[vast shard]

oh ok

guys

[swift fjord]

you dont see a 1/x^2 ... and think to write the antiderivative in that form lol

[vast shard]

this is what the solution suggested

[swift fjord]

what i said

😎

[twilit spindle]

that is not what I would've done

[twilit spindle]

chatgpt*

[swift fjord]

https://tenor.com/view/stephen-curry-nba-legends-golden-state-warriors-steph-curry-gif-26163672

[vast shard]

also i have a general question. both q1 and q9 are 2 functions being multiplied together. im just wondering how to know if i should be using integration by parts versus u-sub ||i asked deepseek and it said i have to look if one part is the derivative of the other, is it trustable|| do yall have any tips? cuz i would be confused whether i should use integration by parts or not

[twilit spindle]

in general when we see multiplication IBP is the way to go

beccause notably int udv=uv-intvdu

[swift fjord]

bro said notably [the definition for integration by parts]

also surely u check u sub first

[vast shard]

uhh

[swift fjord]

always see if theres a u-sub approach first

[twilit spindle]

I said notably then gave the formula

[swift fjord]

if thers not (should be quick to check) consider by parts

[vast shard]

how exactky should i check

[twilit spindle]

if it matches the form int f'(g(x))g'(x)dx

[swift fjord]

see if theres any term thats a derivative of another term

in the first example you can see that theres cosx and sinx

so you can think if u replaced sinx with u

[twilit spindle]

since d/dx(f(g(x))=f'(g(x))g'(x) ==> f(g(x))=int f'(g(x))g'(x)dx

[swift fjord]

would that be easy to solve

integral of 1 + u^2 is very easy

done

[vast shard]

ohhhhhhhhhhhhhhhh

okok thanks both of yall

[twilit spindle]

IBP should be obvious, if you see product of two weird functions that don't relate in derivatives it's not a bad bet

[vast shard]

oke

[twilit spindle]

but with integration there's not rule set in stone on what rule you should apply like derivatives

[vast shard]

another question then, how do i know what to set as u and what to set as dv

because i think it makes a difference

[vast shard]

trial and error 🥀

[twilit spindle]

LIATE is a good rule of thumb

set u in order:
log
inverse
algebraic
trig
exponential

[vast shard]

is that real

i thoguht it was smth chatgpt made up

LOLLLL

[twilit spindle]

😭

just don't overrely on AI

[vast shard]

yaya

surely yall are more helpfuk

helpful and reliable

🔥

[twilit spindle]

well it's that but also thinking critically and longly about a problem to gain udnerstanding is important

[vast shard]

yep frfr

[safe radish]

@vast shard Has your question been resolved?

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[molten acorn]

!status

[safe radish]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[molten acorn]

Do you know what formulas to use?

[graceful lichen]

the formulas from kinematics of circular motion?

[graceful lichen]

No

Are you here @molten acorn

[molten acorn]

Yes

Sorry for dissapearing

[graceful lichen]

I believe intuitively a_c >= f_max

[molten acorn]

The coin slips when the maximum friction = centripetal force

[graceful lichen]

This should be condition if it slips?

[graceful lichen]

It also can be greater right?

[molten acorn]

It says 'just slips', so we should set = instead of >=

Do you understand why?

[graceful lichen]

no

Yes got it

It had just started to slip so that the a_c = f_max

[molten acorn]

Yep

[graceful lichen]

At that moment it will be =

[molten acorn]

Now you can plug in the formulas into that

You can ping me if you get stuck

[graceful lichen]

mw²r = umg

w²r= ug

4w²(40) = u(10)

[molten acorn]

You are disrespecting my pfp by writing mu as u :p

But alright lol

[graceful lichen]

Haha

16w² = u

Now?

[molten acorn]

Then, we know m(2w)^2 r_2 = umg

wait sorry

[graceful lichen]

I used question condition here

[molten acorn]

r_2 is the radius when the coin slips when the speed is doubled

[molten acorn]

Yah

This one is different

[molten acorn]

Now you need to solve for r_2

[graceful lichen]

I am getting 10cm

[molten acorn]

Yep

[graceful lichen]

Yes

[molten acorn]

Well done

[graceful lichen]

.close

[safe radish]

Channel closed

Closed by @graceful lichen

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[graceful lichen]

.reopen

[safe radish]

✅ Original question: #help-23 message

[molten acorn]

What happened

[graceful lichen]

I have one more question

Well I solved it rn

I am getting 36

I thought I couldn't solve it back when I tried so I asked but when I asked and read question I got correct answer

Lol

.close

[safe radish]

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[safe radish]

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[graceful lichen]

Q4

The normal acceleration in the question is the acceleration due to gravity or is it the acceleration of the body "a"

[safe radish]

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[safe radish]

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[bleak lark]

how did (2n-1)!/(2n+1)(2n)(2n-1)! happen?

[astral glacier]

What's n!

Like what's the expression for it

[bleak lark]

um

what do you mean

[astral glacier]

What is 5!

[bleak lark]

5x4x3x2x1

[astral glacier]

Good

Now what's 4!

[bleak lark]

4x3x2x1

[astral glacier]

Good

Now can you write 5! in terms of 4!

[bleak lark]

5 x 4! ?

[astral glacier]

Yes

So now what's (2n+1)! in terms of (2n)!

Also hi Green's theorem

[bleak lark]

(2n+1)! = (2n+1) (2n+1-1) = (2n+1) (2n)!

oh and (2n)! = (2n) (2n-1)!

i see

[astral glacier]

Yes

[bleak lark]

.close

[safe radish]

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[safe radish]

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[compact merlin]

What is happening here?!?

[compact merlin]

This makes no sense

Usually if the summation starts at 1 or something then you do k-1 =n and n=k+1

This thing is making it 2k for some reason

I've never seen a problem with a sub n+2 in the recurrence relation

Is it treating the 2k and 2k+1 as your a0 and a1 "solutions"

[dusty raven]

I'm guessing these recurrence relations come from the Taylor expansion of a solution to the ODE.

I'm not sure what you mean by

"Is it treating the 2k and 2k+1 as your a0 and a1 "solutions""

but essentially it is just solving the recurrence relation. It splits it into odd (2k) and even (2k+1) indices since these behave independently (e.g. a_{2k+1} only depends on the previous odd number index which is a_{2k-1} and so on)

[compact merlin]

Yes but usually the goal is to find y1 and y2

where y1 has a0 in it and y2 has a1 in it

and then it magically turns into series form which I do not know how to do

[dusty raven]

Are you happy with how to solve the recurrence relation?

How to use it to solve the ODE is a separate question (which we can discuss)

[compact merlin]

Well I know we set all the coefficients equal to 0

But in class it was always just like solving for a2,a3,a4,a5, etc in terms of a0 and a1

But I think it uses 2k because its a sub n+2 instead of just an

[dusty raven]

the coefficients of what? This is probably correct but it would help to make sure we're talking about the same thing

[dusty raven]

yes this is correct

(that is, it uses 2k and 2k+1)

[compact merlin]

The differential equation when you do y and y' and y''

You move stuff around so all the summations start at the same point

Then solve for stuff using the coefficients of x^n

[dusty raven]

yes

You're assuming the solution takes the form $y=\sum_{n=0}^{\infty}a_nx^n$ and then deriving conditions those coefficients must satisfy

[flat frigate]

LayneTheAbomination

[compact merlin]

Yeah so I can't do the last step

Without messing it up in some way

[dusty raven]

What is the last step?

[compact merlin]

When you have the recurrence relation you are supposed to plug in n = 0,1,2,3...

and then magically get a series or something

idk how to do that even without k involved

[dusty raven]

Hmm, you already do have a series though (the sequence $a_n$ is determined up to the undetermined constants $a_2$ and $a_0$)

[dusty raven]

could you give me an example of what you mean here?

[flat frigate]

LayneTheAbomination

[compact merlin]

well I know a2 = -a0

But that doesn't help anything

We were just supposed to solve that one because it was in the differential equation

but the important stuff is in the summation

[dusty raven]

I'm still not clear what you mean by the "last step"

Do you want to write down the solution to this differential equation (as a series? in closed form if possible?)

[compact merlin]

Yes

The last step is announcing your y1 and y2

Like you would for a normal person level differential equation where you just factor r^2-2r-3 or something

The series ones are so stupid

[dusty raven]

The series method is very useful but a bit more tedious yes

But anyway, okay, can you have a guess at what $y_1$ would be (as a series)? Hint: there's a natural way to split up your series based on the fact that the odd terms all involve $a_1$ and the even terms all involve $a_0$

[flat frigate]

LayneTheAbomination

[compact merlin]

y1 =summation 1 to infinity of (-1)^n * x^2n/(2n)!

[dusty raven]

I'm going to put what you wrote in LaTeX just so we can check it easier

$y_1=\sum_{n=1}^\infty \frac{(-1)^nx^{2n}}{(2n)!}$

[flat frigate]

LayneTheAbomination

[dusty raven]

this is close, but note that $(2n-1)\cdot{}(2n-3)...5\cdot{}3\cdot{}1\neq{}(2n-1)!$

[flat frigate]

LayneTheAbomination

[compact merlin]

-1 in the bottom factorial for odd

2n-1

[dusty raven]

it's actually the other way around if you look back at your original message, e.g., $a_{2k}=\frac{(-1)^k}{(2k-1)\cdot{}...\cdot{}5\cdot{}3\cdot{1}}$

[flat frigate]

LayneTheAbomination

[compact merlin]

So what am I supposed to do

you mean 2n+1 factorial?

[dusty raven]

First off, do we agree on what the series is?

And no, I don't mean (2n+1) factorial

[dusty raven]

like I write here, the product of the first k odd number is not equal a factorial

[compact merlin]

There is like always a factorial in these things

because taylor series

I thought

[dusty raven]

You could rewrite it it to involve a factorial

And there's a special notation for these products off odd or even numbers (called "double factorials")

But the point more is that once you've written down the series then either you

a) are done if there's no closed form expression
b) just need to look a list of Taylor series and see if yours matches

[compact merlin]

I don't have a list in my head

I haven't done calc 2 in like 2 years

by closed form do you mean no summation

[dusty raven]

I also don't have a list in my head, which is why I suggested looking it up

[dusty raven]

No I mean like

$\sum_{n=0}^\infty{}\frac{x^n}{n!}=e^x$

[flat frigate]

LayneTheAbomination

[compact merlin]

Ok I just did another sample problem and I see what I am doing now

Well I still don't know how to put it in series form

[dusty raven]

Anyway I think we can agree:

$y_1=\sum_{k=0}^{\infty}a_{2k}x^{2k}$

and

$y_2=\sum_{k=0}^{\infty}a_{2k+1}x^{2k+1}$.

Next, you also identified that you just need to (correctly) substitute the expressions for $a_2k$ and $a_{2k+1}$ into the above equations for $y_1$ and $y_2$

[compact merlin]

But I know how to find terms from the recurrence relation

[dusty raven]

great!

[flat frigate]

LayneTheAbomination

[compact merlin]

Can I just keep it as a sub n+2 and plug in 0,1,2,...

instead of doing k

[dusty raven]

you can use $n$ instead of $k$ sure

[flat frigate]

LayneTheAbomination

[dusty raven]

what exactly do you mean here? if you do an example I can verify (but we've basically answered the question. you only need to substitute the expressions for a_{2k} and a_{2k+1})

[compact merlin]

and then put it in summation form

I'm bad at seeing the patterns

[dusty raven]

what i've written here is in summation form, isn't it?

[compact merlin]

Yes

Mkay

those are the solutions

[dusty raven]

yes. they aren't complete since you have expressions for a_{2k} and a_{2k+1}, but those are the solutions

[safe radish]

@compact merlin Has your question been resolved?

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[shadow cloud]

hi

I need help with this

[safe radish]

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[shadow cloud]

i need help

[burnt notch]

What is your question?

[shadow cloud]

i dont know how to solve 4x^2-5x-12=0

[burnt notch]

There's plenty of ways to do it

It really depends on what you have been taught

By the way, the one in the photo is one of such methods

[shadow cloud]

i am used to completing square

[shadow cloud]

yeah my home teacher taught me

but i dont understand

i asked him earlier he said that u need to let all of it divided by 4 and do (a-b)^2

[burnt notch]

Mmhh

[shadow cloud]

please help

[jagged belfry]

it might be useful to try to solve for the general equation ax^2+bx+c = 0 (a \neq 0)

[shadow cloud]

change all the + to -

[jagged belfry]

sure. it wouldnt matter

[shadow cloud]

please help

is this how we do it?

[jagged belfry]

suppose ax^2+bx+c = 0

then x^2 + b/a x + c/a = 0

[shadow cloud]

can u screenshare in paint

[jagged belfry]

what is (x+c)(x+d)?

[shadow cloud]

what.

[jagged belfry]

you want to find c such that (x+c)^2 = x^2 + b/a x + h where h is some constant

[shadow cloud]

wq8

w8

[flint vessel]

the method is correct theres an error in your calculation tho