help-27

[misty crest]

well yea? do you understand this

if you understand this then it should be clear

[hazy cypress]

yes, but i don't get how it relates to my question. it seems like an unrelated fact

[misty crest]

it’s not at all unrelated

[hazy cypress]

i can understand that, i just don't understand how

[misty crest]

we first compute lim |a_{n + 1}/a_n| and after simplification we found that the limit equals
|x - 2|

do you agree with this so far?

[hazy cypress]

yes

[misty crest]

ok and the ratio test tells us that if this limit is < 1 then the series for a_n converges

since we are interested in where our series converge we should therefore set our limit < 1 to see what values of x make that inequality true

doing so leads us to |x - 2| < 1

we rewrite this as
-1 < x - 2 < 1 <—> 1 < x < 3

so we know for sure that the series converges in this interval then we check the endpoints to see if it converges there as well

you’re wondering why the radius is 1?

the radius is the distance from the center

the center here is 2

|x - 2| < 1 means all points within a distance 1 from 2

youll notice that for any power series we have some |x - c| term after our limit computation

and the limit will always be of the form L * |x - c|

we want this to be < 1 so we get
|x - c| < 1/L

1/L would be our radius

note that if L = 0 then
L * |x - c| < 1 always holds since 0 < 1 regardless of the value of x so our radius would be infinite in a sense

and if L = inf then L * |x - c| < 1 would only hold if x = c

every power series converges at its center (since we’re just summing zeros)

[hazy cypress]

sorry, gimme sec

got sidetracked

i think i get it

[misty crest]

🔥

[late void]

she hugged me and I got tingles from it

[misty crest]

interesting

[hazy cypress]

.close

[devout snow]

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[hushed vault]

How do I do number 5? I dunno what they mean by secant line 😭 😭

[uncut crow]

the line that goes through the 2 points

[young spade]

find a line of the form y = mx + b
that passes through the points (x1, f(x1)) and (x2,f(x2))

[hushed vault]

...how do i do that

sorry

[young spade]

In the previous problem you already solved what is the value of "m"

[hushed vault]

yeah for the slope

[young spade]

Now you only have to find what is b

For that you can solve a simple equation

[hushed vault]

do i have to do point slope form?

[young spade]

You can do that too, but its arguably easier to just find the value of b based on some of the two values of x1 or x2

$$f(x_1) = m x_1 + b$$
$$f(x_1) - m x_1 = b$$

[young spade]

And you already know what f(x1), m and x1 are.

you could do the same with x2, and prob you would be inclined to check with both just in case, both should give the same value for b.

[hushed vault]

that makes a lot more sense

tysm

i have another question aswell

[young spade]

go ahead

[hushed vault]

in the picture i included it showed a table at the bottom, for number 9

freak no

sorry number 6

the table is for problem no. 6

and questions 8 and 9 are:

8. Where is the rate of change constant?

9. Write the equation that represents the average rate of change

[young spade]

To be really honest, youre probably a few classes away from knowing about derivatives and these problems are more suited for it. But we can do it with the avg change of rate too

[hushed vault]

ngl, this is my final review packet 😭

[young spade]

oh, lmao.

[hushed vault]

yeah rip

[young spade]

Well, anyways, dont worry about it. We will work under the assumption that the avg rate of change is taken in ranges of length 1

[hushed vault]

mkay

[young spade]

I hope you see that all ranges go from a to a+1

[hushed vault]

yeah

[young spade]

Now, for 8, we would like a range that as a avg. rate of change of 0

[hushed vault]

wait sorry, what does that mean

im honestly so sorry, im kinda sick right now and all of this is a different language, i swear im smarter when unsick 😭

[young spade]

To be really honest, the question is a little ambiguous on words. I wont bore you on the why, but this function doesnt really have any place where the rate of change becomes actually constant.

[hushed vault]

for context, this is a first semester ap precalc packet if that helps?

[young spade]

But im mostly working in that they meant for a place in the function in where the avg rate of change is 0, aka, it says as is in the bounds of a range.

[hushed vault]

thats what im sayinnng, i literally dunno what they mean

[hushed vault]

yeah that would make sense

[young spade]

in another words, we would like the following:

remembering that all ranges go from a to a+1

$\frac{f(a+1)-f(a)}{(a+1) - a} = 0$

[young spade]

I wrote it with parenthesis to make it obvious, but the denominator obviously simplifies to 1

And since the only way a fraction is = 0 is that the numerator is also = 0, then we can ignore the whole denominator part even if it wasnt 1.

[hushed vault]

okay, im gonna ask my teacher on monday just incase but i think you have it right so far

[young spade]

after simplification, we end up with

f(a+1) = f(a), you agree?

[hushed vault]

yeah sounds right

[young spade]

knowing that our f(x) is 4 - x^2, you should be able to find a value of a that satisfies the equation.

remembering that here, a is the value that determines the range [a,a+1]

[hushed vault]

yeah

okay perf

sorry, oneee last question

[young spade]

go ahead

[hushed vault]

freak

sorry hold on

[young spade]

do a screenshot prob

[hushed vault]

there ya go

this is the one, i just dunno how to like, do this bro 😭 😭

[young spade]

Ooh i was misreading this one and was banging my head so hard on it.

Well

[hushed vault]

PFFT

me too bro me too 💔

[young spade]

Well, first

Any given polynomial of degree "n" has n roots too.

Open to multiplicity

[hushed vault]

yeah i'm ngl i don't understand

[young spade]

Do you remember whats the multiplicity of a root?

aka a "zero" of the function

[hushed vault]

yeah, the power right?

[young spade]

kind of

[hushed vault]

the biggest power

[young spade]

Nope, thats the degree of the polynomial

[hushed vault]

whoops

[young spade]

for example, x^4 - 4x^3 + 6x^2 - 4x + 1 is a degree 4 polynomial

[hushed vault]

yeah cause 4 is the biggest power

right?

[young spade]

yeah

There is a really important theorem literally called the Fundamental Theorem of Algebra

I wont bore you with the full detail

[hushed vault]

guys why did my teacher not teach me this 💔

[young spade]

But the tl;dr is that a "n-degree" polynomial will have n amount of roots. > down to multiplicity

So, here, this polynomial will have 4 roots.

[hushed vault]

oh yeahh, i remember that, we just did not get the name or anything

[young spade]

And just in case, the roots might be either real or complex.

[hushed vault]

yeah cause -3i is one of the roots

[young spade]

in this particular case (cause i wrote the polynomial myself) this only has 1 type of root.

x = 1

but this root has multiplicity 4

thats because, if we factorize the polynomial, we get (x-1)^4

[hushed vault]

ohhhh

so its the power the root is to when factored down

that makes alot more sense

so it would be like (x-4)^2

[young spade]

And you should remember that any n-degree polynomial can also be factorized into n-factors

Now, we want a 4 degree polynomial, so we are missing two factors yet

[hushed vault]

so 4 factors

[young spade]

One of them is already given to us, -3i

So we have:
$(x-4)^2(x+3i)(x-a)$ where a is our missing factor/root

[young spade]

You will agree that (x-4)^2 is entirely real.

[hushed vault]

would the third factor not just be (x-3i)?

ohh okay

perfecto, thank you so much

[young spade]

Yep, you appeal to diff of squares to argue that you get no complex coefficients

depending on the problem, they will prob ask you to expand the factorization

Which, yeah, takes a bit.

[hushed vault]

alrighty, that twas my last question, tysm!

[young spade]

np, good luck

[hushed vault]

.close

[devout snow]

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[devout snow]

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[limpid lance]

This quesiton js tricky

[limpid lance]

For (c), I was able to obtain N11.9E however AI got a different answer

[sand quarry]

,rccw

[crystal dawn]

I'm not sure if I'm in the right headspace to understand this rn, but any time you have a discrepency with another source, please show the full working of both you and the source

[sand quarry]

This question has a lot of steps you need to take, so please do what Hanako said and show your working

[devout snow]

@limpid lance Has your question been resolved?

[devout snow]

@limpid lance Has your question been resolved?

[limpid lance]

N12.9E sorry I forgot

AI’s answers

Let’s compute it cleanly and give the numerical bearing.

---

1. Resolve velocities into components

Take:

• East = +x
• North = +y

Athlete A: 3 m/s at N30°W

[
\vec v_A =
\begin{cases}
x_A = -3\sin30^\circ = -1.5
y_A = 3\cos30^\circ = 2.598
\end{cases}
]

Athlete B: 2 m/s at N10°E

[
\vec v_B =
\begin{cases}
x_B = 2\sin10^\circ = 0.347
y_B = 2\cos10^\circ = 1.970
\end{cases}
]

---

2. Relative velocity of B with respect to A

[
\vec v_{B/A} = \vec v_B - \vec v_A
]

[
\vec v_{B/A} =
\begin{cases}
x = 0.347 - (-1.5) = 1.847
y = 1.970 - 2.598 = -0.628
\end{cases}
]

So relative velocity points E and S.

---

3. Direction of separation at closest approach

At closest approach:
[
\vec r_{BA} \perp \vec v_{B/A}
]

Thus, the direction of (\vec r_{BA}) has slope:
[
\frac{y}{x} = -\frac{x_{v}}{y_{v}} = -\frac{1.847}{-0.628} = 2.94
]

So:
[
\tan\theta = \frac{x}{y} = \frac{1}{2.94}
\Rightarrow \theta \approx 18.8^\circ
]

This angle is measured from North towards East.

---

4. Direction athlete A must take to intercept B

Athlete A must run along the line joining A to B, i.e. along (\vec r_{BA}).

[
\boxed{\text{Direction} = \textbf{N }19^\circ\textbf{ E}}
]

---

✅ Final Answer

[
\boxed{\text{Athlete A should run in the direction } \textbf{N }19^\circ\textbf{ E}}
]

If you want, I can also show:

• A labelled vector diagram, or
• A short exam-marking version (3–4 lines).

[woven radish]

S
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[limpid lance]

Different models keep getting different answers

Gemini got 11.9

[limpid lance]

I have provided my full working, can you help please?

[devout snow]

@limpid lance Has your question been resolved?

[devout snow]

@limpid lance Has your question been resolved?

[pseudo basin]

!noai

[devout snow]

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).

[young spade]

I hope this helps.

And i also hope i read it correctly.

@limpid lance

[devout snow]

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[cinder tendon]

Is this correct way to show? i havent done one in so long

[lavish sigil]

yep

[cinder tendon]

so even the arrow is right?

[slate peak]

It's an inconsequential detail, no?

[lavish sigil]

I mean sure? But I'd personally not format it that way

[uneven coral]

I write it more like a => than a ->, but oh well, I don't think it matters that much here

[lavish sigil]

,tex \begin{align*}
\int_0^2 (x + 1)^{-1/2}\dd{x} &= \int_1^3 u^{-1/2}\dd{u} \tag{$u = x + 1$} \
&= \frac{u^{1/2}}{1/2}\eval_1^3 = 2\sqrt{3} - 2 \approx 1.464
\end{align*}

Can TeXit not do align environments lol

[woven radish]

jewels!

[lavish sigil]

(Looks better on a document)

[devout snow]

@cinder tendon Has your question been resolved?

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[cinder tendon]

thanks

[cinder tendon]

.reopen sorry, i wrote it like this . i believe its correct now but just wanna double check

[devout snow]

✅ Original question: #help-27 message

[lunar harbor]

looks good

[devout snow]

@cinder tendon Has your question been resolved?

[devout snow]

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[placid cove]

Why is the chain rule being used here? Am I misunderstanding theorem 2?

[versed juniper]

chain rule is always there

that's the du/dx

[placid cove]

Does it have something to do with the du/dx in part b? I get pretty lost with that notation

i see

[uncut crow]

the ‘u’ there is 2x^2 + 1

[placid cove]

oh I see is it like d/dx * u = du/dx? maybe im dumb but i feel like that could have been more clear

.close

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[hard sable]

not sure about

[hard sable]

aii

i know that p - 3x >= 0

as graph is in the 1st quadrant

[drifting sierra]

Just plug in the point (3, 1/2)

[hard sable]

yeah so p - 9 >= 0, p >= 9

wait

you mean into the equation?

[drifting sierra]

Yes

[hard sable]

bruh i was overcomplicating it

.close

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[spiral burrow]

how can i find the coordinates of the red dot

[native garnet]

Are there any measurements or is this more of a proof?

[spiral burrow]

the radius is 50cm

wait i have some more

[pseudo basin]

!xy

[devout snow]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[spiral burrow]

the angle that the y line makes with the blue line is 22,5

english is not my native language

[pseudo basin]

do not call me "bro".

[willow helm]

Yeah it wasn't about that

[pseudo basin]

also yeah it wasnt about language

it's that you gave us only pieces of the problem and not the full thing

can we have a picture of the full problem? we need to be sure you didn't miss anything else.

[spiral burrow]

im tryna design a centrifugal

for a jet engine

i have to put some weir angly things

i need the cordinates to do that

[native garnet]

Did you find the problem in a textbook/worksheet?

[spiral burrow]

no

i tried to do it like this 300x=22,5.360

im kinda dumb

[willow helm]

What's 22,5.360

[spiral burrow]

The angle that the y-axis makes with the dotted blue line. is 22,5

i tried to do it like that if 360 degress equals to 300 whats 22,5 equals to

[willow helm]

uh

what?

[spiral burrow]

yeah i have to ask this to my teacher

[willow helm]

360 degrees equals 300

[spiral burrow]

the circumference of the circle is 300 degrees

3.2.50

[willow helm]

What's your language

[spiral burrow]

turkish

[willow helm]

Can you post everything in turkish

Because the translations don't do much good I am afraid

[spiral burrow]

çemberin yarıçapı 50cm ve kırmızı bir nokta var bu nokta bu çemberin 22,5 derecelik kısmına denk gelen şeklin ucunda yer alıyor pi sayısını 3 olarak alacağız kırmızı noktanın kordinatları lazım

[native garnet]

I tried redrawing the graph based on the information given.

[willow helm]

The radius of the circle is 50cm, and there is a red dot at the end of the shape that corresponds to the 22.5-degree part of this circle. We will assume the value of pi as 3. The coordinates of the red dot are needed. (I guess this is the translation)

||you lost me at assuming pi = 3 :(||

[zinc flame]

"assume the value of pi as 3" yup, definitely building a jet engine alright

[spiral burrow]

im just tryna find the right solution also im not even an university student what do you expect me to do

[zinc flame]

I was joking, don't worry about it. anyway you're on the right track

[native garnet]

Yes, this feels like a geometry/trigonometry problem to me.

[spiral burrow]

ty i kinda lied im not designing a jet engine but this is a assigment that my teacher gave me

[spiral burrow]

what do i do

:(

.close

[devout snow]

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[willow helm]

hmm

have you learned about polar coordinates?

[spiral burrow]

.reopen

[devout snow]

✅ Original question: #help-27 message

[spiral burrow]

i guess

i have to translate that

this thing?

[willow helm]

Yes

[spiral burrow]

imma go to sleep i just stole yall time sorry

not intentionaly

i swear

[willow helm]

No problem

good night

[spiral burrow]

thanks

im so sorry

please forgive me

[willow helm]

don't worry about it

[spiral burrow]

i mustve knew english before i came here

:(

.close

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[delicate fulcrum]

I’m a second-year student at an Italian scientific high school.
Our math teacher gave our class this geometric riddle to solve. Almost nobody really tried to work it out, but I solved it using similar triangles.
At the end of the lesson, I showed my solution to the teacher, and he asked me how I managed to find the result without explicitly using the horizontal distance between P1 and P2.
The point is that I did use it implicitly, but it cancels out because of triangle similarity.
From similar triangles, I wrote:
x / 4 = (6 - x) / 6
Solving step by step:
6x = 4(6 - x)
6x = 24 - 4x
10x = 24
x = 2.4
At first I was confused because swapping the terms gives 6 - x = 3.6, but that is just the complementary height, not a different solution.
I believe the solution is correct, but I would really appreciate it if someone could confirm it and explain clearly why this method works.

[versed juniper]

the triangles arent similar though

they only share a side

nothing else

[uneven coral]

also the right angle (looks like it)

[versed juniper]

oh are you talking about the one formed by the person in the middle

[delicate fulcrum]

There is a right angle (as stated by the teacher), and the two triangles also share the angle at the intersection, so they have two equal angles and are therefore similar.

[versed juniper]

icwym

[uneven coral]

ahhhhhhh

do you mean the CP1O and DP2O triangle (O being his head)

[versed juniper]

those arent similar

[uneven coral]

yeah because the right angle is no longer present

but in any other triangle I don't see the equal angle at intersection

[versed juniper]

doing it algebraically with equations of the lines on a graph, the distance between the heights cancel out

[delicate fulcrum]

So, the only thing that wasn’t clear to me was why, if I swap the denominators, I get one value in one way, 2.4, and another value, 3.6, in the other way. And if I then subtract 3.6 from 6, I get 2.4. That’s the only thing I didn’t understand

[uneven coral]

ig what if we extend the 4 to 6, the connect it to D and extend x to that

maybe then geometrically it'd make some sense

[versed juniper]

wdym swap denominators

[delicate fulcrum]

Here’s a photo of my calculations and how I tried to solve it

[uneven coral]

god I don't want to make a diagram, but I kinda get it

It's basically like if you extend the x to as high as 6

and x could be either the x we have, or the 6 - x above it

then the value of x you'd get is 2.4 and 3.6, which would depend on which distance you found (the height of the man, or the extra distance above him till it's 6 metres)

[delicate fulcrum]

ooh ok so that’s how it is, i get it now, if this is the solution, thanks, didn’t get how the two proportions worked before, sorry if i stressed you out

[uneven coral]

IG it's almost like, the ratio between the 4 and the height of the man, is equal to the ratio between 6, and the height above that man till 6 metres

I can't exactly prove how though, although I do kinda see how that might work

[delicate fulcrum]

yeah i got that, actually i realized it right away when i saw the exercise, that’s why i did the proportion, it’s just that i didn’t understand why putting the denominator one way gave one result and swapping it gave another, thanks a lot

[uneven coral]

Ahhhh no problem man, happy to help

[delicate fulcrum]

Thanks !

[devout snow]

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[visual stone]

During a collision between two objects, momentum is conserved if there is no loss of energy. Thus pt = pAi + pBi = pAf + pBf , in which pt corresponds to the total momentum (in kg·m/s) and ( pAi, pBi, pAf, pBf) represent the initial and final momenta (in kg x m/s) of objets A and B. Even more, p = mv (p and v are vectors) in which p corresponds to the momentum (in kg x m/s) of an object, (m) its mass (in kg), and (v) its velocity (in m/s).

The diagram illustrates a curling throw, where stone A, with a mass of 18.8 kg, collides with stone B, with a mass of 17.5 kg.

What is the magnitude and direction of the velocity of stone A after the collision?

[visual stone]

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[turbid imp]

Super quick question, for euler's and fermat's theorem in discrete maths, is the y != 0 always, or just for positive whole numbers, etc.?

[turbid imp]

so as in, i don't know how to define y in it, should it be "y is a positive integer", y € N, y != 0, y € Z_+, or something else?

(and is it the same across both euler and fermat*)

[vital edge]

What exactly do you mean by Euler's and Fermat's theorems

There's too many of them

[vagrant skiff]

The most precise definition is y in Z and gcd(y,n)=1

or gcd(y,p) = 1 in Fermat’s

[turbid imp]

this page says it's positive integer for both

https://math.libretexts.org/Bookshelves/Combinatorics_and_Discrete_Mathematics/Elementary_Number_Theory_(Raji)/03%3A_Congruences/3.05%3A_Theorems_of_Fermat_Euler_and_Wilson

my lectures (swedish tbf, but still*) are saying for fermat's it's y != 0 and im not even sure for euler's

[vital edge]

Ah little theorem

You take a prime, and smth that the prime doesn't divide

Which, by the nature of primes, means that the gcd is 1 as cooly said

[turbid imp]

so y€Z?

[vagrant skiff]

U_m is the group of units Modulo m

So

glow$U_m = {, y in \Z_m \mid gcd(y,m) = 1,}$

That automatically excludes y = 0

[vital edge]

Why is your texit gay

How do I convert mine

[vagrant skiff]

Simply pay jet 100 bucks a day

[woven radish]

クーリー

[turbid imp]

so can I just say y != 0 and then sgd(y,m)=1 => Y^phi(m) =_m 1? (where the = is actually the triple equalsign for kongruent)

wikipedia says "any integer a" (a is y in their case) for fermat's and for euler's "a positive integer"

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[potent compass]

Please help

[crystal dawn]

combine the 2s in the numerator and denom separately first

have you tried this at least? ^

[potent compass]

I’m doing it now

Tysm

Is it always lien this though?

[crystal dawn]

wdym

[potent compass]

Uh Ig I get another question similar to this

Do I do the same thing

[crystal dawn]

I mean, yes? I don't see why not unless you wanna show the question

after all, all I see here is multiplication and division, and they are commutative

[potent compass]

Ty

[lunar harbor]

orz u

[potent compass]

Can I multiply the final answer?

[shut trail]

yes

powers are the same

so the bases can multiply

[potent compass]

Ohh tysm

Ty

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[steep quartz]

is the teacher wrong or what

[lunar harbor]

,w increasing -x^3-3x^2+5

[woven radish]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by civil_service_pigeon

[steep quartz]

its increasing as long as x is negative

oh

bruh

wait i factored wrong

lol

mb

[lunar harbor]

[steep quartz]

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[solid osprey]

any ideas on how to find primes $p_1,p_2,p_3$ (not nessicarily distinct) such that
$$p_1+p_2+p_3+7=p_1p_2p_3$$

[woven radish]

ihave<skissue>

[solid osprey]

i mean a cubic (RHS) should grow much faster than somethinng linear (LHS), but how can i formalize that idea?

[hollow ice]

parity seems more promising imo

[solid osprey]

mod 6? i tried that and it became pretty confusing

maybe im just bad

[hollow ice]

nahh

like 3 odd primes + 7 is even

but the product is odd

[solid osprey]

ohh

[hollow ice]

same with one 2 and 2 odds

[solid osprey]

so either all of them is 2 (fails) or one of them is 2

[hollow ice]

so that leaves you with 2x 2

[solid osprey]

wlog its p_3, so p_1+p_2+9=2p_1p_2

[hollow ice]

nah, one being 2 makes the sum odd and product even

[solid osprey]

yeah? whats wrong with that

you will get sum+odd=product
odd+odd=even right?

[hollow ice]

odd = even is wrong

[hollow ice]

what? if only one of them is 2, then p1 + p2 would be even

and even + odd is odd

[solid osprey]

ok im actually crazy

so we want 2 of them to be 2

[hollow ice]

yea

[solid osprey]

so p_1+4+7=4p_1
11/3=p_1

this isnt an integer, so no sols?

[hollow ice]

ig

yea

[solid osprey]

okay thank you!!

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[proven oriole]

Trying to get good practice in, while I understand how to solve radical equations that are like. By themselves (an example being the square root of 19+X is equal to 1)

However when you combine two radicals I start to struggle. Here’s my work and truth be told I’m just a little uncertain if I’m even progressing the first step correctly

[north roost]

${(a+b)^2 \neq a^2 + b^2}$

[woven radish]

k

[north roost]

js for simplification, id use u = 3-2x, maybe?

[fossil locust]

yeah, it would be easier to solve $\sqrt{2u + 1} + \sqrt u = 1$

[woven radish]

south

[proven oriole]

How did you guys transform the problem into that

[fossil locust]

use this and you'll see

[proven oriole]

and I'm assuming this is where I messed up originally? I did A^2+b^2 instead of (a+b)^2?

[fossil locust]

yes, and many other mistakes

if you don't mind I'll list them out one by one

[proven oriole]

Also is u just filling in the variable for X?

[fossil locust]

we define u = 3 - 2x

then you want to get 7 - 4x in terms of u

[proven oriole]

Yes please do. I don't mean to be a lot, but I can't understand what to do right. if I don't know where my misunderstanding starts

[restive river]

hi is this the help channel for high school math?

[fossil locust]

1. first line should be 1 - sqrt(3 - 2x)
2. even if you forget the middle term when squaring [1 - sqrt(3 - 2x)], you should get 1 + (3 - 2x)

when you square root something then square it, you get back the original expression

[fossil locust]

!occupied

[devout snow]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[restive river]

can I help?

[fossil locust]

sure thing

[restive river]

alroghty

[proven oriole]

i just realized i wrote the problem down wrong

okay okay and the reason I'm doing 1-square root of 3-2x is because thats the easiest thing to get rid of?

[restive river]

wait I can write down the sol to explain

[fossil locust]

yes, it's easier if you start squaring from $\sqrt{7 - 4x} = 1 - \sqrt{3 - 2x}$

[woven radish]

south

[fossil locust]

don't give them the full solution though!

otherwise they won't learn anything

[proven oriole]

^^

South gets it

[restive river]

where did they specifically go wrong

[fossil locust]

think about what your middle term would be if you didn't rearrange it first

you'd get $2 \sqrt{7 - 4x} \sqrt{3 - 2x}$

[woven radish]

south

[fossil locust]

@restive river

[restive river]

noted. I'm volunteering to teach a kid so I need to learn a few ways to do so 😭

[proven oriole]

i want it to be known that I'm still here, i'm just rewriting the problem and putting notes between each step. That way you can guys can not only see my work but track my thought process

[restive river]

hey so, south's method will get you there. but in general, when you're solving two radical terms, you must and must get rid of the radical by squaring the RHS and LHS. then you use simple algebra to move ahead

[proven oriole]

okay wait, sorry. I'm looking at this problem of

[steel sage]

If you let y=3-2x, it becomes sqrt(y+1)+sqrt(y)=1. Furthermore you let y=(tan(t))^2 you will obtain (1+tan(t/2))^2=1-(tan(t/2))^2

[restive river]

did you forget the middle term while expanding (a+b)²?

[steel sage]

Maybe overcomplicated but first thing entered my mind

[fossil locust]

letting y = tan^2 t..

[restive river]

@proven oriole are you familiar with trigonometry?

[proven oriole]

no

[steel sage]

Then forget what I said

[proven oriole]

and now theres like 3 people all saying a variety of things

[restive river]

okay, then go with the remove the radical at any cost method

[restive river]

lemme try your way

[north roost]

r we serious...

[restive river]

where'd sqrt(y+1) come from

[steel sage]

sqrt(1+2y), I was so dumb

[proven oriole]

So if I put in this format. Not a lot changes in my head about how to attack the problem. It’s reorder but for what? Now I’ve done a similar problem where there was a number outside of the square root sign

[steel sage]

Anyway y=0 seems an obvious solution

[proven oriole]

Like this, and that’s how I handled it for the correct answer

So I’m a little lost on what the point of moving this one is supposed to be. And how I’m supposed to eliminate it

[steel sage]

sqrt(1+2y)+sqrt(y)=1

[fossil locust]

your original problem has right-hand side 1 though

[fossil locust]

but okay, let's start with this easier question first

[restive river]

get rid of the radical

[fossil locust]

if you have 2y - 1 = 3, then surely 2y = 4

the idea is to have one side be a constant

after $\sqrt[3] {2x + 6} = 4$, cubing both sides gives $2x + 6 = 64$ as you've written down

[woven radish]

south

[proven oriole]

right, i follow this. But why can't I just simply cube both sides for my original problem as well?

or does moving the 1 to the left, make me able to cube?

[fossil locust]

why would you cube both sides when you have square roots?

[proven oriole]

If you cube both sides then you cancel out the square roots. Which would take the problem from Sqrt of 7-4x + sq 3-2x=-1 down to

7-4x+ 3-2x=1

[fossil locust]

no

to undo cube roots, you cube both sides
to undo square roots, you square both sides

[proven oriole]

i'll write that down

[north roost]

\textbf{cubing} ${x}$' means you are doing ${x^3}$, whereas \textbf{squaring} ${x}$' means you are doing ${x^2}$

[woven radish]

k

[proven oriole]

so in this case instead of cubing I'd be squaring both sides

so the problem would look like

Oh

[north roost]

,rccw

[north roost]

you must square

[proven oriole]

Sorry

[north roost]

gng

not cube

like this

[proven oriole]

Shit

Yeah

Yeah

My fault

I said it in my head and wrote down something diffe

[north roost]

${(\sqrt{7-4x})^2 = (1-\sqrt{3-2x})^2}$

[woven radish]

k

[north roost]

this is what u envision, ya?

[proven oriole]

Mhm. In my head square is 2 because it’s 2 dimensional and cube is 3 because it’s 3

I just have to make sure I put the correct numbers when I do this

I apologize again for being slow to pick this up

[rain summit]

Where's the domain restriction 😠

[proven oriole]

What is a domain restriction?

[rain summit]

Domain restriction of $\sqrt x$ is $x \geq 0$

[woven radish]

1 divided by 0 equals Infinity

[rain summit]

Basically values that makes the function defined

[restive river]

yes, here's a quick tip why. \textbf{squaring} ${x}$' means you are multiplying ${x}$ with itself, thus the exponential representation becomes ${x^2}$, the same way the square root of ${x}$ becomes ${x^(1/2)}$ and the CUBE root of x becomes ${x^(1/3)}$ so when you square ${x^(1/2)}$ exponentially the 1/2 is multiplied with 2, which gives 1 and you end up with ${x}$ which eases your expression by a 100!

[woven radish]

smithereens

[restive river]

omg why won't it type 😭

[steel sage]

My way btw, y=tan^2(θ)/2, missed ^2, and in the end =-2sqrt(2) contradiction, I always make lots of errors

[proven oriole]

I’m not understanding where I need to define or input this. I understand what it’s saying I don’t understand how to relate it to my problem

[proven oriole]

Saving this to write down later

[rain summit]

In some problems, you might need this to eliminate some values of $x$

[woven radish]

1 divided by 0 equals Infinity

[fossil locust]

that only applies when you find the solutions to x

there will be a value of x which makes 3 - 2x or 7 - 4x be negative, and you can't take the square root of a negative number

[fossil locust]

you really should expand both sides first

[restive river]

what I mean by x^1/3 is

same goes for sqrt

[restive river]

lemme try

[proven oriole]

I think I got it?

Also the notes on the bottom is for something entirely different ignore that

[fossil locust]

no, you're still making the same $a^2 + b^2$ mistake

[woven radish]

south

[fossil locust]

where's the middle term 2 * 1 * -sqrt(3 - 2x)

and why are you dividing everything by 2

[restive river]

where'd 1+2y coem from? if we're to write ${7-4x}$ in terms of ${3-2x}$ it'll be $sqrt{y+4-2x}$

[proven oriole]

What’s the difference between these two functions? Because if (X+Y)^2
Don’t I have to take everything inside the parenthesis to the second power anyway?

[woven radish]

smithereens

[fossil locust]

$(x + y)^2 = x^2 + 2xy + y^2$

[steel sage]

7-4x=1+2(3-2x)

[woven radish]

south

[restive river]

what? how so?

[proven oriole]

To try and isolate X. I divided 2x by 2 and what you do on one side you have to do to the other. 4x/2x is 2x. Then I repeated it to get X alone

[restive river]

nvm got it

[rain summit]

Did he really squared the whole thing?

[proven oriole]

Going to erase and back up

[restive river]

@steel sage that's very smart 😭 I kinda need help with high school math these days, could I dm for help?

[steel sage]

Sure

[restive river]

is ${y=(tan{theta})/2}$ an assumption?

[woven radish]

smithereens

[proven oriole]

I really like this visual but I’m having trouble understanding it. So

And I’m going to drag this question away from everything else for a moment

[rain summit]

You do know you are in other person's channel

[rain summit]

Do another question first

[restive river]

dude, do you know binomial into binomial multiplication

[restive river]

it's the same question tho

[rain summit]

Still, someone's in a channel for an algebra question, and you are doing a trig one, i suggest opening a new channel and then linking your old convo into the new channel, or continue the convo there

[steel sage]

Sorry, you are right

[restive river]

my bad

[restive river]

hey could you make the other channel? I'm kinda persistent on understanding this

[proven oriole]

How do I apply that middle part of 2XY

That’s the part that’s messing with my brain

[restive river]

do yk binomial and binomial multiplication?

[proven oriole]

Uhm, gimme an example problem of binomial and binomial multiplication

I don’t know what that is by name alone

[restive river]

for eg
${(1+x)(2+x)}$ as such

[woven radish]

smithereens

[restive river]

@proven oriole

[fast garden]

xy is x * y, you are almost there
also, you forgot a minus in the middle

[proven oriole]

I’m answering it, so you can see how I’m going about tackling it

[fossil locust]

you can use the distributive law on this

[fossil locust]

so that would be $1(2 + x) + x(2 + x)$

[woven radish]

south

[fossil locust]

it's like how (1 + x) * 5 = 1 * 5 + x * 5

remember if you want to use FOIL, you should get 4 terms

well FOIL works cause of the distributive law

[proven oriole]

This is how I’d answer that question off my memory. But I feel like I’m neglecting a X (sorry I had to blur out mail)

[proven oriole]

OHHH

TEAH

[rain summit]

where's $x^2$?

[proven oriole]

OKAY

[woven radish]

1 divided by 0 equals Infinity

[restive river]

this implies that ${1(2+x)+x(2+x)}$ the same way
${(a+b)^2}$ is ${(a+b)(a+b)}$

[proven oriole]

Like I said I thought I was missing something with a X

[woven radish]

smithereens

[proven oriole]

How does this multiplication thing tie into the division? Or the A^2+2ab+b^2

[rain summit]

simplifying in fractions

like how $\frac{(a + b)^2}{a + b} = a + b$

[woven radish]

1 divided by 0 equals Infinity

[rain summit]

fractions are basically division

[fossil locust]

you don't need to divide

[restive river]

when you expand,

[proven oriole]

So are you saying

instead of doing A^2 +2AB+B^2
I could do A (A+B) + B(a+b)

In my case it would look like

[rain summit]

if you expand it fully, it's stil $A^2 + 2AB + B^2$

[woven radish]

1 divided by 0 equals Infinity

[restive river]

nope, that was an explanation on why ${(a+b)^2)}$ is ${a^2 + 2ab + b^2}$

[woven radish]

smithereens

[proven oriole]

?

[rain summit]

expand further

[proven oriole]

Oh

[fossil locust]

nearly

7(7 - 4x) - 4x(7 - 4x)

[restive river]

@proven oriole are you understanding why we squared it?

[proven oriole]

I assume it’s -4x because the 4x is negative

And when I took it out for some reason I made it positive

[restive river]

hon idt you're grasping the basics of this

[proven oriole]

I’m not understanding why, but I’m understanding what is being asked of me when I see A^2+2ab+b^2. And I understand a format that I know how to solve

[proven oriole]

Like I can solve this

That I know

[fossil locust]

okay but to simplify $(\sqrt{7 - 4x})^2$ you don't need to do that at all

[restive river]

if you want to expand ${(7-4x)^2}$ then your first step is ${(7-4x)(7-4x)}$. tell me why.

[woven radish]

south

smithereens

[proven oriole]

I need to do (7-4x) (7-4x) because that's that squaring is. I'm doing multiplying that equation twice. Is that answer too simple am I missing something?

[restive river]

idt they're quite grasping the why behind it. you can't rely on a method to solve always if you don't know why you're using it in the first place

[proven oriole]

if i wanted to expand (7-4x)^5 I would do (7-4x) 5 times

[restive river]

yes, good. it is because ${x^2}$ is $x$ times $x$. you got that. now tell me when you have two terms i.e; 7, $4x$, what happens?

[woven radish]

smithereens

[proven oriole]

i'm not understanding what's being asked here. When I have two terms, what happens? What happens when what? What am I doing with the two terms? Two terms on their own don't mean a lot.

[restive river]

(^ represents exponent btw not multiplication), you're almost there! in this, you're doing 5 times of 7 and 5 times of 4x and subtracting it, right

[proven oriole]

thats what i meant

sorry

[restive river]

I'm saying $(7-4x)(7-4x)$, here you have two terms unlike 1 term in the eg ${5(7-4x)}$. so now that yk what to do with the latter, tell me how you will solve the former

[woven radish]

smithereens

[restive river]

did you get my question

@proven oriole ?

which grade are you in btw

I'll have to leave in a few mins @proven oriole :((
@fossil locust take over

[proven oriole]

Yes, and I'm just solving the equation. I don't want you to think I don't know how to do things that I do know, because I struggle to word/explain math. Then if i show a picture of me doing the math you'll know if I understand or not.

and then I'd answer that equation

[restive river]

$(16x^2)$ *

[woven radish]

smithereens

[restive river]

the last term

[proven oriole]

12th im trying to graduate

OH

yep and i understand why it's squared

and i understand why its positive

i rushed

[restive river]

in which country?

[proven oriole]

america?

[restive river]

ahh

the edu system is different here

12th is the last year

for us

idk how it works there lol

[proven oriole]

its the last year for highschool here too

[restive river]

what

and then you're off for college?

[proven oriole]

yeah..?

[restive river]

no more school years?

[restive river]

whattttt

😭

that's the math we did in 6th grade here 😭

im in 10th rn

[proven oriole]

ok well

[restive river]

I can't find someone who's good with their stats :((

[proven oriole]

im happy you got it really early on, you can make fun of the dumb american sterotype in your off time. Please dont do it in front of my face

[restive river]

@fossil locust are you?

[fossil locust]

can you not ping me

I don't want to be involved

[restive river]

oh I'm sorry

[fossil locust]

you can just close this channel, open another one and show your working out

[proven oriole]

alright, thank you for the help

how do you close?

[deep abyss]

.close

[proven oriole]

.close

[devout snow]

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[flat willow]

i have a quistion

how to divide (2x^3 - 4x^2 + 4) by (x-2)

[pseudo basin]

do you know how to do polynomial long division in general? Y/N

[flat willow]

yes

Y

[pseudo basin]

ok, what's troubling you with this one

[winter torrent]

start by writing it as 2x^3 - 4x^2 + 0x + 4

[flat willow]

i did that

[winter torrent]

$\polylongdiv[stage=1]{2x^3 - 4x^2 + 4}{x-2}$

[woven radish]

two wuggen in a trenchcoat

[flat willow]

after the first step, 2X^3 is cancled, but -4X^2 is also cancled

[winter torrent]

sure

[flat willow]

im stuck here

[pseudo basin]

ok then just keep going

[winter torrent]

$\polylongdiv[stage=3]{2x^3 - 4x^2 + 4}{x-2}$

[woven radish]

two wuggen in a trenchcoat

[pseudo basin]

your goal at each step is to kill the highest-degree term @flat willow

if more terms get killed, all the better, right?

you just move on to whichever one would be coming next

[flat willow]

how did you got positive 4X^2 ?

[pseudo basin]

(well, except all that remains would be 4. and you're kinda just done)

[pseudo basin]

-(2x^3 - 4x^2) = -2x^3 + 4x^2.

[flat willow]

how did you do that?

[pseudo basin]

do what, exactly...?

all i did was explain how @winter torrent wrote down her first step of the polynomial division

which should result in much the same thing as what you did

[flat willow]

multiplying 2x^3 - 4x^2 by -1

[rain summit]

how did you get that $\LaTeX$?

i tried in $\LaTeX$ testing but i can't do it

[flat willow]

?????

[winter torrent]

well... we're subtracting it, which is the same as multiplying by -1 and then adding

it's just easier to read that way

[pseudo basin]

do you mean:

1. "Why are we multiplying 2x^3-4x^2 by -1 at all?"
2. "I get why this would be done, but I don't understand the execution of it."

[flat willow]

but when we mulitiply 2x^2 by -2 we get -4X^2

-4X^2 - (-4X^2) = 0

[pseudo basin]

correct

[devout snow]

@flat willow Has your question been resolved?

[devout snow]

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[flat willow]

yes, Thanks

[devout snow]

@flat willow Has your question been resolved?

[flat willow]

yes

thank\

[winter torrent]

.close

[devout snow]

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[proven oriole]

I don’t understand where I’m supposed to go from here.

In case my hand writing isn’t clear enough. The problem is The square root of 8x+1 is equal to x+2

[drifting sierra]

$\sqrt{8x+1} = x+2$, find x?

[woven radish]

Nel

[proven oriole]

Yes I’m trying to find X

[sand pumice]

line 8 you should have 0 on LHS

[proven oriole]

What does LHS Stand for?

[drifting sierra]

From 7 to 8, you did -8x

[sand pumice]

left hand side

i mean on the left side of the = sign

[drifting sierra]

8x - 8x is not x

[proven oriole]

Gotcha

Sorry in my head since “x” is undefined it might as well equal Zero. So I don’t ever remove X because it could be literally anything

[drifting sierra]

That's not at all how you should think about it

x is not undefined, it's unknown (for now)

Saying it might as well equal zero is just incorrect

[proven oriole]

What’s the difference between undefined and unknown, idk if that’s a dumb question

[quartz spindle]

hii im new

[drifting sierra]

1/0 is undefined

So is 0/0, 0^0, or just "infinity"

Those are not numbers

x is a variable, it is supposed to have a value

Now if the expression you're working on leads to x = 0/0 or something like that, you can conclude that x is undefined

But before that, x is simply just unknown: you don't know its value

[proven oriole]

Got it, I never had someone correct me on calling X that before

[meager vale]

3 , 1 ?

[proven oriole]

But I can see the difference between 1/0 which isn’t possible or infinity oppose to unknown

[drifting sierra]

Right... first things first: your first step was squaring both sides, but that only keeps the equivalence if both sides have the same sign

So before you do that step, you should at the very least state what range of value x can take

You do know that sqrt(x) is only defined for x >= 0, right?

[proven oriole]

I did not, no. So how would writing that down look like?

[drifting sierra]

Writing what down exactly?

[proven oriole]

This range thing

[drifting sierra]

Just use words

$\sqrt{x}$ is only defined for $x \ge 0$

[woven radish]

Nel

[drifting sierra]

That's perfectly fine to write

Now in your case it's sqrt(8x+1), so for what range of values of x is that defined?

[proven oriole]

Sorry I saw you typing and didn’t want to interrupt, but the square root of x must be greater than or equal to one in this case..? Is that a proper range? I don’t think it could be 8 because whatever X is must be multiplied by 8 but regardless of its outcome it will be added by one

If it’s being added by one it can’t be zero

[drifting sierra]

You're not quite thinking about it correctly

Whatever is inside the sqrt must be greater than or equal to zero

That means 8x+1 >= 0

Can you solve this inequality?

[proven oriole]

Oh, I overthought that

But yes you can solve that

[quartz spindle]

sorry to interrupt you but i think that x may be 3 or 1

im in 10th grade and we have been taught about this

[drifting sierra]

You're not the helpee

!nosols

[devout snow]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[proven oriole]

I don’t want the answer, I want to know how to get there

[quartz spindle]

oh

sorry i did not knew that

[proven oriole]

It’s okay, you didn’t know lol

[quartz spindle]

actually, you can square both sides and then use Discriminant formula

use discriminant formula to further solve it

[drifting sierra]

Are you solving it? I'm not sure what you're waiting for here

[quartz spindle]

btw in which grade do u study

[proven oriole]

Oh, sorry I blanked out for a moment. The answer to that is -1/8 which makes the equation not true

[quartz spindle]

what?

[drifting sierra]

Ok but you should write the full thing

By just saying -1/8 nobody knows whether you mean x = -1/8 or something else

In this case it's x >= -1/8, right?

[quartz spindle]

yo the value of x is either 3 or 1

ig i am correct

[drifting sierra]

Could you stop interrupting?

[proven oriole]

Sorry, yes the statement
8x+1>=0 is untrue because X equals -1/8’s

[drifting sierra]

No, that's still incorrect

[quartz spindle]

what the heck is this server for?

[proven oriole]

Dog just read all info channels

[drifting sierra]

Why are you saying "8x+1>=0 is untrue [...]"? That's not what solving means

[proven oriole]

I don’t know what you mean by solving, I assumed you were asking me to break down the equation further. So I broke down 8x+1 then found out it couldn’t be right because if it simplifies to a negative it’s below zero

[drifting sierra]

But 8x+1 >= 0 is clearly true if x is 2, for example

Let's take a step back

You are given the equation $\sqrt{8x+1} = x+2$, and asked to find its solutions, that is the set of values for x such that the equality holds

[woven radish]

Nel

[drifting sierra]

Let's try a few values for x and see what happens

We'll try 0, 1, and -1

[proven oriole]

I can do that, want me to write it out and show you how I do it?

[drifting sierra]

Sure go ahead

[proven oriole]

Awesome give me a moment as you gave me 3 numbers to punch in

[quartz spindle]

Hello

If you want then I can help you

In simple way

?

Or I am going to sleep

[proven oriole]

Alright I don’t know how I’d punch in zero tbh. I also don’t know how this relates to what the original problem is asking

But I did it

[proven oriole]

Go ahead and sleep, it’s okay 👍

[quartz spindle]

What ur answer

Tell

Me

[drifting sierra]

Hold on, you wrote sqrt(8 - 1 + 1) ?

[proven oriole]

Yeah

[drifting sierra]

The expression is sqrt(8x+1), not sqrt(8+x+1)

[proven oriole]

Well im putting -1 in for x right?

[drifting sierra]

8 and x are multiplied together

[proven oriole]

Ohh that explains why I struggled to get a similar concept

[drifting sierra]

This is very basic algebra, no wonder you're struggling with this exercise

When you have an expression and you want to plug in a specific value for a variable, you need to think of it as replacing the variable with that value in parentheses

So sqrt(8x+1) becomes sqrt(8(-1)+1)

8 and -1 are still being multiplied

[proven oriole]

I thought about that but second guessed it and decided to just go super literal. I’m a little rusty as I haven’t touched math in quite some time

Just remembering a lot of rules to actually do things

Okay I can do that though

[drifting sierra]

Alright, try again

You should know what to do with 0 too, now

[proven oriole]

Yep

Quick question sorry

[proven oriole]

Sqrt 8(-1)+1 but would it also be Sqrt 8(-1)+1 =(-1)+2?

[drifting sierra]

Yes, the other side is (-1)+2, but here the parentheses are less useful

[proven oriole]

Okay sorry for the pause

Obviously the last one is also false

Also to be super clear, I didn’t wanna take up too much paper space. So since it was real simple multiplication I then added the one each time

Oh I forgot to put on the square root over them all

[drifting sierra]

Ok but the sqrt disappeared

[proven oriole]

Yeah I noticed that as I was staring at it

The square root of 9 is 3

[drifting sierra]

Also be careful what the sqrt contains, here it looks like the +1 is not included

[proven oriole]

I’ll draw the lines thicker so I don’t confuse you or my future instructor

My fault

[drifting sierra]

Well it also doesn't help that the line is on top of the paper guide line

Anyway, sqrt(9) does indeed equal 3

[proven oriole]

Yeah, but uhm 1 is true 0 and -1 are not.

[drifting sierra]

Right, for x=0 you get sqrt(1) which is just 1, but the other side is 2, so the equality does not hold

[proven oriole]

While that’s a good skill to have in my tool box it doesn’t exactly inform me how to continue this equation. It doesn’t however REALLY help me understand what and why I’m doing this

I don’t want you to think I’m sassing you or nothin

[proven oriole]

Mhm! 📝

[drifting sierra]

You're missing fundamental knowledge so I'm trying to go over it quickly so that solving the actual exercise makes sense

[proven oriole]

I appreciate that greatly

[drifting sierra]

Now, when x = -1, can you tell me what happens?

[proven oriole]

Idk how to word this

It turns the 8 into a negative.

[drifting sierra]

Yes, and you end up with sqrt(-7)

What does this mean?

Is sqrt(-7) something you can work with?

[proven oriole]

No you can’t, there are no whole numbers that like square root to -7

If that’s what you’re asking

[drifting sierra]

Right

So we say sqrt(-7) is undefined

As soon as anything in an equation is undefined, you can stop and say the equation doesn't hold

[proven oriole]

OHHH

[drifting sierra]

That's where -1/8 comes in

[proven oriole]

wait that’s so sick

[drifting sierra]

sqrt(8x+1) is only defined when 8x+1 >= 0, so when x >= -1/8

Do you understand that?

[proven oriole]

Yes

YES

[drifting sierra]

Now you can start solving $\sqrt{8x+1} = x+2$

[woven radish]

Nel

[drifting sierra]

You can square both sides, but you must keep in mind that:

• x >= -1/8, otherwise one side is undefined so the equation no longer holds
• both sides must have the same sign (you can go back and check this later by plugging in your solutions into the original equation)

For the second point, in this case, you can also just say that the left side is always >= 0 since it's a square root, and so x+2 must also be >= 0

That means x >= -2

But you already have x >= -1/8 so this other condition x >= -2 doesn't matter

So far so good?

[proven oriole]

Yep I’m following along

[drifting sierra]

So here you got to 8x = x^2 + 4x + 3, which is correct

Then you tried to subtract 8x but did it wrong

[proven oriole]

Because I said 8x-8x=x instead of 0
Or did I misunderstand 4x-8x?

[drifting sierra]

The former, yes

[lunar harbor]

hi

[drifting sierra]

Hi

[proven oriole]

Gotcha

Gotcha gotcha gotcha

But even when I do it properly and 0=x^2-4x+3

How would I break this down further?

Wait actually

I can take that 3 away

Because it doesn’t have a X on it

[drifting sierra]

No, you can't

[proven oriole]

Dang it

[drifting sierra]

If you do that you need to take 3 away from the 0 as well

[proven oriole]

That was the plan

Because when you do stuff like that it goes on both sides

[drifting sierra]

So -3 = x^2 - 4x ?

[proven oriole]

Mhm!

[drifting sierra]

Ok but that doesn't really help

This is the point where you need to know how to solve quadratic equations

There are many ways to do that, one systematic way is simply to apply the quadratic formula

Another way is to "complete the square"

Yet another is to factorize, which is a lot easier when you already know one solution

[proven oriole]

What’s the quadratic formula? Formula’s are pretty simple once you remember which one you need to attach to what

[drifting sierra]

,, x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

[woven radish]

Nel

[proven oriole]

Okay wow

Hm

[drifting sierra]

For any equation ax^2 + bx + c = 0, this gives you the x values that satisfy it

[proven oriole]

That’s a +- sign right? Meaning it can be either?

[drifting sierra]

Yes, there are at most two solutions

If the thing in the square root is negative, there are no solutions, if it's zero, there is exactly one, since the plus-or-minus doesn't matter

If it's positive, then you get two, one for the plus, one for the minus

[proven oriole]

Gotcha

[drifting sierra]

That thing, b^2 - 4ac, is called the discriminant

(it discriminates between the number of solutions: 0, 1, 2)

[proven oriole]

So beforehand you gave me really
Simple questions for me to break down. Can you do the same with this quadractic equation? That way I know how to properly plug in numbers into it

I can also then put it down on my note page of quadratic equations

[drifting sierra]

Well you have x^2 - 4x + 3 = 0

You need to identify what a, b, and c are

Such that if you plug them back into ax^2 + bx + c = 0, you get your equation back

[proven oriole]

Okay so

[proven oriole]

Sorry let me finish writing this part down and I’m I’ll try to plug in what’s what

Now if this formula makes any sort of sense. Then A B C is the exact order the numbers appear in. Which is also why it’s so important to rewrite my equation from left to right. Something I didn’t do until now lol

Which makes me question is the X part in the A B C formula literally supposed to be X in the equation or am I missing something?

[drifting sierra]

Yes it is

[proven oriole]

So in that case

[proven oriole]

Okay so that means in order to find X. I need to plug in the letters I just found into this here.

[drifting sierra]

Yes

Make sure your a,b,c values are correct

[proven oriole]

Now before I go through the effort of painstakingly do the math. Do I need to convert numbers over as is? What I mean by that is:

for example. B for me is -4x. But according to the Quadratic formula B must be negative. Do I need to make my -4x positive since I’m mashing together two negatives? What if my B just so happened to be 4x? Would I make it negative to match the equation?

Secondary question. It messes with my brain when formulas do stuff like “4ac” or “2a” but thanks to you I know that when formulas put variables tightly together like that they need to be multiplied right? With paranthesis around them at that

[drifting sierra]

Ok first of all, a,b,c do not contain x. They are the values you can plug into ax^2 + bx + c = 0 to get your equation x^2 - 4x + 3 = 0

[proven oriole]

Oh that’s weird but I get it

[drifting sierra]

If a was x^2 like you tried to do, you would get (x^2)x^2 + ... = 0, which is not the same as your equation x^2 + ... = 0

Then, you say "b must be negative", but that's not the case at all

Are you saying that just because the formula starts with "-b" ?

[proven oriole]

Mhm

[drifting sierra]

b can be any real number

[proven oriole]

I was asking in fear of taking this far too literally

[drifting sierra]

-b is just that number, multiplied by -1

[proven oriole]

So when plugging in -4, because I don’t bring down the X. I’d be doing -4*-1 to place it in that first slot

That would make it a positive 4

[drifting sierra]

Yes, if b = -4, then -b = -(-4) = 4

[proven oriole]

Okay I’m getting it I’m getting it

I can clean this up

[proven oriole]

Is my “A” (x^2) an exception to this rule or not bringing X’s into the equation? If not then what would I put in its place

[drifting sierra]

You want ax^2 = x^2

What value of a satisfies that?

[proven oriole]

Typing out my thought process here
I want Ax^2 to =X^2
But x is unknown. My gut reaction says if I want something to not change I plug in 1 because 1 copies whatever it’s raised to or multiplied.

But 2 (which is 1^2) ≠ X^2

So I just put in X by itself and drop the exponent??

Because X^2=x^2

[drifting sierra]

But 2 (which is 1^2) ≠ X^2
Not sure what you mean

[proven oriole]

Okay so I want Ax^2=X^2

If I plug in 1x^2

Oh

Sorry I think I’m just wrong

I was going to say 1x^2 wouldn’t equal X^2

But I mistakingly did 1^2 instead of just leaving it at 1x^2

[drifting sierra]

Right

So a = 1

[proven oriole]

Am I finally ready to solve this?

[drifting sierra]

c = 0 ?

[proven oriole]

How did I get there

Whoops

[drifting sierra]

Also, -4^2 is not the same as (-4)^2, careful

[proven oriole]

Wow I need to parenthesis everything

Okay yeah, i got that

[drifting sierra]

You do, but once you get used to it, you can skip a lot

For example you really don't need to write the (1) factors, since they don't change anything (in a product)

[proven oriole]

I’ll get to that level after a few drills of solving radicals right now I gotta stay to the basics 💪

Okay I’m not gonna send another picture of the same equation but! I did parenthesis every single thing I plugged in

This is the easiest part, now I’m not trying to find out what I’m solving. Now I’m doing the fun part of actually solving lol

If you add you get positive 3

And subtract you get !

1

@drifting sierra holy hell. I got the answer right. Here’s my work for the addition the subtraction is on another page. (As you can see I ran out of space)

But! I want to say you were great

And thank you so much for dedicating so much time to me. Not once did you make me feel bad about not knowing this or like stupid or anything

[drifting sierra]

Good job

[proven oriole]

Please have an amazing day i feel so pumped now

.close