I differentiated the quadratic formula
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- Опубликовано: 1 окт 2024
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This is my first time differentiating the quadratic formula and I did it wrong first! We can first solve the quadratic equation ax^2+bx+c=0 and will see that x is actually a function of a, b, and c. So when we differentiate the quadratic equation with respect to a, either we have to do partial derivative or we will have to keep in mind that x is a function of a and use implicit differentiation.
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What you need is d(roots) / d(coefficients) which is possible, because there is ONE MORE COEFFICIENT than there are ROOTS. This derivative is actually a matrix of coefficients.
If you try d(coefficients) / d(roots) it doesn't work. It might work if you reduce it to a standard form, for example the highest power of x could have the coefficient 1. Then you would be differentiating the coefficients wrt THE SAME NUMBER of roots.
As Shakespeare wrote, 2 b squared or not 2 b squared: that is the question!
@@simonmultiverse6349 I think you slighlty misquoted the great mathematician, Shakespear. I believe it was, "2b squared or negative 2b squared: THAT is the question!"
i have a question. can we differentiate that?
since ‘ax^2+bx+c=0’ is not an identity but an equation, i think we cant differentiate that expression.
@@최민영-g2d Yes, because you differentiate each of the terms. Since it's an equation, the derivative is also an equation. For example if you have P+Q=R+S then the derivative would be dP+dQ = dR+dS
"It was wrong because it was too easy" A wise monk - some year
That's how I always knew my homework was wrong. :D
@@herbie_the_hillbillie_goat 😂
@@herbie_the_hillbillie_goat XD
It's easy if you are not sweating.
My multivariable Chen Lu senses are tingling haha
😆
Do we have enough Chen Lu and Lu Chen supporters guys? It's Dr P on the horizon!
Hahaha 😂
For those who are confused:
In the function: f(x) = ax^2 + bx + c
x can take any value (independent of a)
Df/da = x^2
In the equation: ax^2 + bx + c = 0
x now refers to the set of solutions x1, x2, which depend on a.
Hope it helps
This looks like what happens when my alphabet soup has broken bits in it.
Thanks for clearing that up, i was really confused at first!
This guy has so much courage ,he could solve such complicated expressions with a smile on his face .
because it’s not his first attempt LOL
What other part of his body would he have a smile!?
S (black hole) dx ;-)
This is not complicated at all for a math bachelor
Nothing complicated about it
Sensitivity analysis is important in engineering because the parameters we use are usually somewhat uncertain. It helps to know what a 5% deviation in a parameter like “a” would do to our answer “x”.
Interesting
Exactly, that's what analysis is good for.
Yes, understanding how to optimize a system (usually a combination of equations) can be very insightful -- but also a complex problem. From my limited experience, optimization studies are only performed when optimization is identified as a necessary objective. Instead, I tried to use these types of parametric studies just to understand how systems actually worked in good detail when others were somewhat arbitrarily choosing their design point that reached at least ONE solution. Due to complexity, just choosing can save a great deal of time so long as you are able to reach a satisfactory outcome. When the outcome was not satisfactory, the habit was to just try again - when optimization studies could instead have pointed to a much more educated choice. It has been some time since I was in college and it still seems that this type of approach is only lightly taught.
Bro, your thinking is just out of the box 📦. You are inspiration for many students. You teach us how to think.
Bro, from which university you studied ?
@@YuvrajSingh-en2zd look his previous video, first sentance. :)
@@toxhydre8806
Ah yes, in his video titled "you, me and the Stanford math tournament algebra tiebreaker", he said he graduated from UC Berkeley as an undergrad. Thank you so much
"Let's do some math for fun" means I drop everything I concentrate on the screen! : )
Thanks! 😆
Seeing him not simplify the fraction by taking out a factor of 2 hurt my soul
I feel it, too!!
But I suppose he ran out of white board space and only wanted to go as far as needed to match up the two different approaches to the derivative.
Still… I have to simplify or I cannot sleep at night! Haha
Its amazing how easy I get distracted from revising my further math exam tomorrow with another math video
Same here hahaha
😅🙏
Sir how to find this integral
int -2^3 abs(x)^[x] dx,
[.]= Greatest integer function
One more integral
int sin(x)^cos(x) dx
This first one is pretty easy - split it into 5 integrals and solve them each individually.
This is an excellent approach that you considered *x* as the function of *a* . But as you know, *x* is also the function of *b* and *c* .
So if *x = f(a, b, c)* then we have:
The derivative of *x* with respect to *a* as you calculate is: --> *δx/δa = -x^2 / (2ax + b)*
So we can calculate also the derivative with respect to *b* as: ---> *δx/δb = -x / (2ax + b)*
And with respect to *c* as: ---> *δx/δc = -1 / (2ax + b)*
That is more like upper-division calculus. It is very hard to grasp at first.
A fascinating pattern emerges: the partial derivatives with respect to x_n (the constant term for x^n) are of the form -x^n / (2ax + b).
those deltas looks super cool, how you do it?! =0
@@roros2512 That's what I'm tryna find out! Use to be able to copy and paste from Wikepedia, but can't anymore.
@@roros2512 those are dirac deltas though? they’re not usually meant to denote partial derivatives….
Your original expression was not partial x/partial a, it was partial(ax^2+bx+c)/partial a = partial 0/partial a = 0. The solution to that is x^2=0, so x=0 as you wrote, ie if you have ax^2+bx+c = 0, changing a continues to make the expression equal zero.
What you solved is certainly more interesting, but your initial formulation of the problem was correct until you stated it implied partial x/partial a=0.
Exactly. The way in which the question was formulated, partial derivatives with respect to any of a,b,c, or x are 0, since changing them will not change the result of the equation which is zero by definition
Indeed. Talking about differentiating an equation is very sloppy - what you are doing is differentiating both sides of an equation. Sloppy language/notation often leads to mistakes like this.
This explanation is actually still a bit wrong. If you have ax^2+bx+c = 0, changing a doesn't usually keep the left-hand side equal to zero. In fact, changing a to a+Δa makes the left-hand side equal to x^2 Δa. What solving the equation partial(ax^2+bx+c)/partial a = x^2 = 0 actually tells you is for what value of x it is true that changing a (by a small amount) leaves ax^2+bx+c unchanged. That happens for x=0, because in that case ax^2+bx+c = c, so changing a has no effect.
The broadest construct views all of the variables as functions of all the others, so b is not constant with respect to a. You can see this world if k of a natural number N as the product of two integers written using the division algorithm as (px+c)(qx+d)=N. The coefficients are going to be related to the particular x as a ring base and Vice versa.
So if x = f(a, b, c) then we have:
The derivative of x with respect to a as you calculate is: --> δx/δa = -x^2 / (2ax + b)
So we can calculate also the derivative with respect to b as: ---> δx/δb = -x / (2ax + b)
And with respect to c as: ---> δx/δc = -1 / (2ax + b)
Hello! And when dx/da=-x^2/(b+2ax)=-(bx+c)/[(b+2ax)*a] here is easer substitute x. or we can even try to solve differential equation :)
And naturaly: NO WAR
Fun stuff! You should look up root locus analysis in control theory; this analytic method was invented in 1948 by Walter Evans to do exactly what you're trying to do (track the change in the roots of a polynomial in response to a change in coefficients). Rather than restricting to derivatives, his method lets you actually look at the locus of all roots as some parameter changes.
Next up: my first time differentiating the cubic formula
Not a very clean looking answer. Anyways, wanted to say that you helped me gain more interest for math. It has led to me discussing math with my teacher on topics that are harder than what we're learning in class. I belive that is something that pushed my final grade up to A even tough I always got B+ on tests
Love this. Can you please prove that if ax^2+2b'x+c=0, then x=(-b'+/-sqrt((b')^2-ac)/a. This is what I call the EVEN quadratic formula and also can u plz differentiate this if possible?
The proof of this is pretty much trivial; just substitute b=(2b') into the normal quadratic formula.
This is novel, unique and ingenious! Absolutely never and would never think of it in this way!
really cool man. Always making my day with cool math concepts
Glad to hear. Thanks.
Nice discussion! But the statement that "x" is a function of "a" is flawed. The zero crossing is a "function" of a (and also b and c), although I wouldn't use the word function but dependent. And therefore the zero crossing itself is an excplicit term namley
0 = ax^2+bx+c
The equation 0 = ax^2+bx+c for given b and c gives a relation between a and x. For certain given domain and range this relation is relation is one-to-one and onto. Then "x" can be given as a function of "a" and "a" can be given as a function of "x". "zero-crossing" more or less suggests that the equation is interpreted as the zero-crossings of a graph of a quadratic. This is a valid interpretation, but there are other interpretations. See Wikipedia "function" for more on functions, implicit functions, multi-valued functions etc.
It seems to me, the first line contains an incorrect statement. There should be not dx/da, but dF/da, where the F = ax^2+bx+c. And if we dont know that the x is a root of the equation (independent parameter) then it will turn out to be true.
And if we know that the x is the root of equation, that we get the derivative of an implicit function.
Yes. Cannot agree more.
The derivitive depends on your setting. If you fix b, c. You get a curve in this vedio. If you fix x, you get a plane, and it was a different story. Without telling anything, it is dangerous to start work just like that.
Nice. This relationship between x and a has implications for factorization of natural numbers.
Sir how to find nth derivative of x^x
You differentiate it n times and look for a pattern.
@@blackpenredpen thanks sir
@@blackpenredpen I tried that but there's no pattern visible to me. I literally spent 8 hours on it.
@@mathevengers1131 nice bro by the way
I have got the solution.
@@SaurabhKumar-jo6dp can you please send me the solution
I saw the thumbnail and went "oh this sounds like fun. I wonder what happens if you integrate it though". So I integrated it. Then I sat there for a few moments looking at what I'd wrought. Wondering why I'd done that. What did I gain? What is it even trying to tell me? I should be doing my comms and signals homework, and yet here I am, integrating the quadratic formula. Why.
😆 I integrated it too
So u r not alone.
I think that ∂(ax²+bx+c)/∂a = x² indeed. In other words, the whole parabola is linear wrt a; as a increases, the parabola widens in direct proportion. What you are doing is seeing how the zeroes of the parabola depend on a, and that is a more complicated question.
1:40-x is a function of a, so it's not a constant.
It's the same as saying c=-ax^2-bx, a function of a...
Very fun idea for a problem, I really enjoyed this!
Thanks!!!
I am going to sit for my country's military acedemy admission test. it is is one of the hardest in my country and this channel really help me a lot solving math quesions😁
I’m confused on why you are writing the partial derivative with respect to a instead of the regular derivative with respect to a. Doesn’t partial derivative imply that all other variables are constant?
I haven't taken a legitimate calculus class in over a decade, since I was an AP Calculus and Physics student in high school. Now, after covid affecting my industry, my plan is to return to university for an engineering degree. Thank you for uploading this video, very useful and informative. If you don't enjoy calculus, you probably hate sudoku and minesweeper lol.
Your derivative of the quadratic formula can still be simplified. There's a square-root on the denominator which can be rationalised (not mandatory, but a preference for most teachers) and 2 can be factored in the numerator, which cancels a 2 in the denominator.
Can you take integral of it next?
Sure thing!
If you take the limit as a goes to 0 of the quadratic formula, you get the (trivial) formula for solving first degree equations
Really? That's so cool!
You can also do that very easily if you apply L'Hospital rule on the quadratic formula (the limit on "a" and consider two cases for positive and negative "b") Only half a sheet of paper...
DUDE, you look 10 years younger without the beard!
I watch your videos when I'm bored, your way of solving problems is so enjoyable.
This is a bit of a small thing, a quibble really, but I think it would be more precise to say you are looking at the total derivative (the usual derivative, not the partial derivative). In the total derivative, we generally assume every other variable has a dependence on the variable of differentiation until proven otherwise -- so, we end up using the product rule on ax^2, and bx. We are assuming here that b is constant, so in the product rule d(bx)/da = (db/da)x + b(dx/da) = b(dx/da), giving the usual constant multiple rule. On the other hand, if we are using partial derivatives it is usual (though of course not universal) to assume every variable is independent of our variable of differentiation, so it would not be appropriate to expand the product rule on ax^2 -- we would treat x as a constant, unless we have explicitly decided otherwise.
The ambiguity is completely removed if we get more explicit about how we are thinking of the lefthand side here. Is it f(a, b, c, x) = ax^2+bx+c? In that case the partial derivative would certainly, unambiguously be just x^2. Are we saying f(a, b, c) = ax^2+bx+c, with an implied dependence of x on a, b, and c defined (variously) as x=(-b+sqrt(b^2-4ac))/2a or as x=(-b-sqrt(b^2-4ac))/2a? Then the method expressed here makes sense, but we have to be extra careful because the derivatives of f may not strictly be functions, as x is multivalued -- as you pointed out, we should really repeat the procedure using the other root to get a complete answer.
As always a good presentation. I just think it's important to be direct and explicit when comparing the partial and total derivatives, since a lot of students get confused / lost in the manipulations. In particular, I have seen many students stumble at precisely the first line here, because they thought the correct answer using partial derivatives (treating x as a constant) was "too easy". I know I didn't understand this distinction until I had been studying calculus for many years already.
The error you make is that you try to differentiate equation while you can only differentiate a function. If you ignore the =0 part x is no longer function of a :)
Right. Under the condition that x is equal to that expression, doesn't the LHS just evaluate to 0?
What’s wrong with differentiating an equation? How about implicit differentiation?
@@karunk7050 good point there bro, I can give you two answers, one simple another slightly more complex - the simple one is: when you do implicit differentiation you notice you have an equation with a function on each side of the = sign and you differentiate each of these functions separately, so really yo do not differentiate the equations, just two functions that happen to be the same and in such a case their derivatives are also the same so the equal sign holds. The second more complex explanation is that when you have an equation that you try to solve, say x+1 = 4, the equal sign here has a significantly different meaning - it does not say the function on the left is always equal to the function on the right, this is plain false. It only states there is some x where the equation holds. If you try to do implicit differentiation here you will get a result 1= 0 which is clearly absurd, the solution is to always understand what = sign means in a context you are operating.
@@arekkrolak6320 ah right, thank you for the great explanation!
Sir, what would you recommend to learn all high school math? There is no school in my area.
Khan Academy.. it's free and does all math from 1st grade through to vector calculus.. you can pick whatever you want. High school math is just algebra, trigonometry, geometry, and pre-calculus in the United States. Statistics and Calculus are offered as options as well.
I would love to see you differentiate the first form with x isolated, but show us how you could keep simplifying it until it becomes the simpler form you got after. Show how they’re both equal, but starting from the complicated one
Why is x a function of a but b and c are not functions of a? You could solve for either b or c and it would be in terms of a
It could. Then we would be doing db/da dx/da and all that.
Omg… we have a lot of combinations dx/dc, da/dx etc 😆
I think because x Is a variable While b and c are not. So b and c can take a value that doesn't depend on the value of a, but the opposite for x
@blackpenredpen mr blackpen red pen i reallly like your sereies especially the five polynomials equation. So please i have a request the request is can you find or give a video of the actual quartic formula which is : ax^4+bx^3+cx^2+dx+e=0 i really NEED that formula for my studies because i study in grade 8 now.
I just love the pokeball, really nails the aesthetic lol
What is the purpose of this exercise? To practice the rules of derivation? The only rational objective would have been to demonstrate a shortcut to solve the quadratic eqn, or to illustrate how to calculate sensitivities, but what you did is a pointless triviality, my friend.
Why do I see Dalai Lama teaching maths
Why can't we solved x, doing differentiate the formula respect to x.
Differentiation of function:
F(x, a, b, c)=ax^2+bx+c with respect to a is x^2. So, you plane to differentiate not a polinom function, but it roots.
All of us at some point have thought of doing many different things to the quadratic equation
How come you do not have any videos showing iteration which has real life uses. For example you need to guess a particle terminal settling velocity to calculate a particle Reynolds number. This particle Reynolds number is used to calculate the drag coefficient, which is used to calculate the particle terminal settling velocity.
Bc I don’t know it all. 😃
@@blackpenredpen lol
The Quad eqn can always be divided by a (a is never zero otherwise the eqn wouldn't be quadratic). In other words, all a's are 1. It's the b's and c's that may be different from 1.
Wow that was great. Now do it with respect to b lol.
From Master Shifu to a Shaolin Monk. Differentiate that nerds.:D
I think that's the first time i see him using the partial derivative simbol xD
I probably would’ve done implicit differentiation first too it’s a lot easier 😂😂
😅🙏
This looks like trolling at first, but it gets more and more interesting as the answer starts to form.
Sorry but aren't you confusing total and partial derivation?
isnt’t it the total derivative? it’s NOT the partial deriv!
But, why?
?
I like to your video so, thank you for your video.
Thank you Black pen red pen for liking my comment
I was about to suggest that you expressed ðx/ða in terms of x, and then you did - cool 👍
Something interesting must be happening, when D = b^2 - 4ac goes towards zero, as a changes ( for given values of b & c ), and the two roots become identical ( a double root ) - and they then vanish or become complex rather, as D < 0 - a socalled "catastrophe" (Thom)
And similar for b & c of course 😉
Makes me wonder what the metric for the space whose basis is unit vectors along a, b and c looks like
This space is the same as the space of quadratic polynomials in x, so a metric doesn't make sense unless we have a way to define distance between quadratic polynomials. But luckily there's a pretty natural choice, which is the root-mean-square distance on some specific interval [x0, x1]: we can define the distance between two polynomials p and q to be ∫ (p(x) - q(x))^2 dx from x=x0 to x1. Given that, you could work out what the metric is. It's probably pretty complicated-looking, but I can tell you without computing it that it will turn out to be the metric of flat 3-dimensional space in disguise, because you can get an orthonormal basis for this space from appropriately scaled and shifted Legendre polynomials.
This is "Brilliant"! 👍
Since the vertex of a parabola is linearly dependent on b, I suspect the zeroes are too
you are a criminal to not cancel out the 2s in the solution
Partial derivatives are not total derivatives.
I didn’t do total derivative.
so, in other words you evaluated the sensitivity of the roots on changes in a parameter.
Ok now do the integral with respect to a 🤣
The most impressive part about this is, the derivative of this demon is done without any mistake and without rubbing anything.
Now differentiate both cubic formulas
Watching this from UK, nostalgia of gcse and Alevel Maths😅😞, But luv the work. After u understand maths, the satisfaction of getting ur answers right when u look through the back of the book is worth it!
It seems like partial differentiation isn’t too different from normal differentiation
ð/ðb (ax^2 + bx + c = 0)
2ax ðx/ðb + (b ðx/ðb + x) + 0 = 0
(3ax + b) ðx/ðb = 0
ðx/ðb = 0
Interesting
what the… i cant tell if youre trolling but this hurts on so many levels
@@wouterfransen9771
Well, since a is a constant in this case, ax + x would be 2ax, right? Maybe not
@@fanamatakecick97 ax+x=x(a+1). and youre treating a partial derivative as a variable like what
@@wouterfransen9771
I am not treating a partial derivative like a variable, idk where you get off saying that
I differentiated (-b±sqrt(b²-4ac))/(2a) directly. It makes obviously the same answer.
Can you tell me who is given this equation.
My first rule of generalized roots of polynomials - nothing ever simplifies
I got frightened only by watching 😥
Cool 😎 👏
How Can you differentiate an EQUATION? Choosing x as a function of a so the quadratic equation is true, what is the derivative of 0=0 with respect to a?
I was wondering the same thing. I think it's the partial derivative of f(x)=... evaluated at f(x) = 0
In the clip the host had probably assumed x to be a function of a, b & c. Reading a chapter on implicit functions and partial differentiation on them in multivariate calculus texts is suggested.
@@sheungmingchoi6804 Well, that would explain the calculation from the second line on, probably, but still the first expression makes no sense to me.
@@WolfgangKais2 The host probably "went over a bit conventionally", writing that line to signal "differentiating both sides of the equation", a common technique of finding derivatives for implicit function. For details, please refer to some texts about "implicit functions", including its differentiation.
Note: you can simplify further by halfing the entire fraction
How t o deferenciat¿
nice
Observation: If you take the quadratic equation and set a=0, then the solution of the equation will be x=-c/b.
Question: If you take the quadratic formula and take the limit as a approaches zero, should the result of the limit also be -c/b? If not, then is there a contradiction here?
Because that's just a linear equation.
This is a great question!
Let's start by thinking visually.
Fix b and c and think of the straight line graph of y=bx+c, with x-intercept at -b/c. For ease of visualisation, assume b>0. Note that graph goes negative to the left of this root (as its gradient b is positive)
Now take a small positive value of a, and add ax² to y to get the graph of y=bx+c+ax² (the same as y=ax²+bx+c, of course), which will be a parabola.
If a is small enough, there will still be a root close to x=-b/c, and the graph will still go negative to the left of this root.
But, because of the ax² term with a>0, the parabola will eventually go off towards infinity as x→-∞, so will eventually cross the x-axis to the left of the root close to -b/c.
We conclude that as a→0 (from above), one root will converge to -b/c, and the other root will tend to -∞.
The behaviour is similar if a
I am so confused!
why didn't we differentiate with respect to anything else?? like x or b or c?
You could if you want to. The a-coefficient is just the example he happened to choose for this example.
Essentially, what he is doing is finding the root sensitivity to the coefficients. A common application of this, is control system theory, where you are interested in how the poles of the transfer function (that are characteristic of the behavior it produces) are sensitive to the physical parameters you can adjust. This is part of the analysis method called the root locus, where we graph the path of the poles, as we vary the gain constant K.
When you did the first equation, why didn’t you factor out a two and cancel with the bottom? Wouldn’t that have simplified the first equation even more?
What do you think about the war between Russia and Ukraine?
Putin needs to leave Ukraine alone, and find a diplomatic solution to any use of Ukraine's land that he desires.
What if x is function of a, b, and c???
It is. That’s why we did the partial derivative
5:33 can't the coefficients be simplified since they're all divisible by 2
He forgot
To find an analytical solution for ax^2+bx+c = 0 is easy. However, what is the analytical solution for ax^(2+e)+bx+c=0 with ‘e’ being a real number? The solutions are
x1=(b/(az))Wq(((-c/b)^z)(a/b)z)^(1/z), where z = (1+e) and q = 1-1/z.
x2 = (-y(a/b)Wq((-1/y)(b/a)((-c/a)^(-1/y))))^(-1/(1+e)) where y = (2+e)/(1+e) and q = 1+y
Wq is the Lambert-Tsallis function (a generalization of the Lambert function). Sometimes the correct solution is x1, in other cases the correct one is x2 and there are cases where x1 = x2, depending on the values of a, b and c. For example the solution of x^(2.5)+x-1 = 0 is x1 = x2 = 0.6540 (up to 4 decimals).
The fury as you left the answer in a reducible form. Divide by 2 on top and bottom... please
Aren’t we differentiating 0(constant) and it will give derivative zero?….help me!
Boa tarde;
X = { -b/2 + - ^[ (-b/2)^2 - (c.a) ] } : a
Obs; quando (b) for n.o ímpar, multiplique a equação por .(2), evitará frações...
x^2 - 5x + 6 = 0 . (2)
2x - 10x + 12 = 0
X = { 5 + - ^[ 25 - 24 ] } : 2
X' = { 5 + 1 } : 2
X' = 3
X" = { 5 - 1 } : 2
X" = 2
Parabéns Mr Black, sucesso sempre.
Brilliant! I didn't think beyond the first method. Thanks. I learned something new today.
The notation I used in this video is very abusive. One should distinct partial derivative and total derivative. Consider f(x,a,b,c)=ax^2+bx+c. Usually the partial derivative (delta f/delta a) means treating “a” as a constant. Indeed, the solution is x^2. But in the second part you interprete x as a function of a,b,c (here, solution of the equation). This derivative should be described as the total derivative of f with respect to a (df/da). The formula used in the second part is exactly how one would compute this derivative (chain rule). More precisely, one should make clear what is meant by differentiating an equation.
-It's not really an equation of 4 variables (a,b,c,x) , it's an equation of 3 variables, since x is just a function of the other 3 (a,b,c)...
-a partial derivative wrt a, b , or c must account for that...
-cool💡...
I was thinking that why did he upload this it's too easy now i get it 😆
Did u think the answer is x^2 too?
@@blackpenredpen yes!
d(ax^2+bx+c)=x^2 da + x db + dc + (2ax + b) dx, so if that is zero and db and dc are zero then dx/da=-(x^2)/(2ax+b). Plug in x matching the quadratic formula and you're done. It's kind of a nice way to go.
Is there a place I can get that derivatives poster/plaque behind you? I absolutely love it!
Yes. It’s in my merch store. Link in description. Thanks.
@@blackpenredpen Thanks! Will be ordering shortly!
@@blackpenredpen I ordered the Calc 2 integrals plaque; super cool! Keep it up!
Now do this with a cubic equation!
♥️🙏
Me(Someone who just learned what a vector was yesterday): Mmmmhhhhmm, yes. The ABC didn't 123, cuz the Sharingan on the left got upset at them.