Can You Pass Harvard's Entrance Exam?

[@prithvichidri9634]

For the guys who directly substituted x= -2🤙

[@ernesthakey3396]

Yep - math nerd with an MIT degree, just looking at it I figured X needs to be negative so that it turns into an addition, -1 doesn't work, -2 does, done. Of course for an exam question they probably require you to show your work, and you don't get credit for just saying "It is intuitively obvious!"

[@vvTejedor]

Yes, but youcan say, that according to Vieta's formulas if there is a rational root, 12 can be divided to it.

[@kutenor]

In my head in -2^3 seconds.

[@arturkowalski5151]

5 seconds for me. And I have studied history, not mathematic...

[@AasimSyedGPlus]

But that's not the only solution

[@markjones5268]

The correct answer is: My parent’s just donated $5M to the school.

[@aliahmed800]

😅

[@AakashSharma-le8pm]

You have to pass the school as well. Donation can only get you admission. ​@@aliahmed800

[@gaynzz6841]

Yup

[@orsolyadr.szepesi8615]

😂😂

[@divermike8943]

@@markjones5268 LOL!

[@wolfhard2420]

Brings me back to Highschool, when the math teacher filled the blackboard with lots of mystery and lacks of explanation skills, so we understood nothing

[@davidb9562]

But this guy "explained" too much during most of the video (extremely tedious) and then rushed it at the end. Big fail.

[@user-tk1ih4zr2s]

just like my teachers, they memorized it but didn't understand it and therefore could only repeat it but not explain it.

[@VaraLaFey]

That is ALL math "teachers" I have EVER seen. They like to show how smart they are, so they restate problems unnecessarily, and suddenly we're not sure what happened. So we try to figure it out - and now we've fallen behind. Which is part of why I, who was skipped a grade in school, have nonetheless been a math-cripple my entire life. That said, I was almost able to solve the initial problem in my head. I suspected x = -2, but in the very last step forgot that 4 - (-8) is 12 instead of -12. Oh well. I'd never get even that far on a more complex problem.

[@creounity]

Why wasn't she explaining enough? That is her direct obligation to do this!

[@CalvinNHobbes1985]

It was pretty clear. Just say your bad at math instead of blaming others for your lack of intelligence

[@ConceivedSorrow]

I solved it this way: x³-x² = 12 Extract common factor x²(1-x) = 12 12 can be written like 2²•3 (prime factorization) so: x²(1-x) = 2²•3 And comparte the factors: x² = 2² 1-x = 3 x= -2

[@kingsircheng4540]

👍

[@johnsmith1953x]

That's only one of the answers=FAIL.

[@johannesvalks]

but it is mathematically incorrect. you miss 2 of the three solutions...

[@lashamez]

Solution only works for integers, not real numbers

[@thebig12conference73]

AND YOU FAILED! x^2 - x^3 = x^2 (1 - x) and 12 = 2^2 x 3 but the problem is that we choose x = 2, we don't have 1-2 = 3. No problem, just replace the 2 in -2 because 2^2 = (-2)^2. To recap, x^2 (1 - x) = (-2)^2 (1 - (-2)) , so -2 is a solution. Then, just factor by x - (-2 ) to get x^2 - 3x + 6 and check that there are no other solutions.

[@divermike8943]

Oh ! He didn't calculate the Complex root! That's not the Harvard Way! The roots are X=-2, X= (3 +sqrt (15) i) / 2, X=(3-sqrt(15) i) /2

[@JimmyD806]

But completing the square would have been a whole other page of math. LOL

[@RexxSchneider]

@@JimmyD806 ... and using the quadratic formula would take maybe two more lines. Worth it to get all three solutions.

[@ericanthonyjones2131]

like finding the three cube roots of 1. including the two complex roots

[@arghyanildey7563]

Here is your admission letter from Harvard. Now give us dollars

[@ObitoUchiha10089]

Exactly my though man. 😂

[@daniellemb1592]

I am very bad at maths, and had always been. I watched at x² - x³ = 12 and thought "x has to be negative, let's try with -2", so I calculated mentally and wrote 4+8=12. Seemed good ! So x= -2 Don't ask for the demonstration ! 😃

[@gerardsmadja1323]

seems obvious , x==2(1-x0 = 12. x should be an integer, and the only factors that can be considered with one of them a perfect square are 4 and 3 so 1 - x = 3 and x = -2

[@rvqx]

@@gerardsmadja1323 Correct, first find one solution and then divide the cubic equation by x minus that solution. So then you must (-x³ +x²-12) divide by (x+2) and that gives you x²-3x+6 which gives you the 2 complex roots. That`s the common way to solve a cubic equation in the math world. In normal world you use a calculator.

[@gerardsmadja1323]

@@rvqx Perfectly right. I assumed wrongly that the iinteger result was the one xpected.

[@steveperry1344]

that's what i came up with also, just a shorter way of doing it and no pencil and paper. i haven't done math in over 50 yearsand only got as far as algerbra ll in high school.

[@einherz]

same. exactly same:)

[@tom-kz9pb]

Seeing -2 only took about 2 seconds. For Harvard, you should need to find the complex solution, as well.

[@potrahsel4195]

Not sure about that.

[@AllDogsAreGoodDogs]

Wolfram came up with x=-2 under "real" solutions. I agree with @potrahsel4195 .

[@chrisclub3185]

Just divide the poly by x + 1 then use the quadratic formula on the resulting quadratic. If this is what’s required to get into Harvard, Harvard has fallen deeply

[@chrisj5443]

Then it should be a harder problem. I'm dumb. It took me 5 seconds.

[@suhumm]

​@@chrisclub3185 by x+2, not by x+1 ))

[@celestine5340]

Different approach: a bit trial and error gives x=-2 as a solution. So dividing the whole equation -x^3 + x^2 -12 = 0 by x + 2 gives -x^2 + 3x - 6 = 0. Which is quadratic, and thus solvable -> but has only imaginairy solutions. I think doing it this way is easier than the method used in the movie.

[@ItsVideos]

Interesting. Thanks. Note that the video method also does a bit of "trial and error" when searching for powers of 2 that sum to 12.

[@christianeb8494]

Exactement.

[@BobJones-rs1sd]

To be a bit more rigorous in finding the first root: one should immediately note that a cubic equation will either have three real roots OR 1 real root and a complex conjugate pair. (Complex roots always come in pairs, which means an odd order polynomial always has at least one real root.) Next, note that the Rational Root Theorem restricts any real rational roots to factors of 12 here. (The Rational Root Theorem states that any rational roots of a polynomial equation are factors of the constant term (here 12) divided by the coefficient of the highest power term (here 1).) Any cubic problem like this on an exam will have a rational real root, as otherwise it would likely require using the general cubic formula to solve with nested cube roots and square roots -- and unless you've specifically studied that formula, no one expects you to memorize it. Inspection of the form tells us to get a positive number on the right side requires that X be negative. So, just start plugging in negative factors of 12: i.e., -1, -2, -3, -4, -6, or -12 are the only possibilities. -1 doesn't work. -2 does work. Done. Divide polynomial by x+2 and proceed as you said. Yes, random guess and check could work in this case since the root is a very small integer. But noting the required type of roots and what possible rational roots exist could narrow down possibilities for more complicated equations where the root isn't an integer or is larger. And would be a lot quicker and less error-prone than the method shown in this video.

[@q-tuber7034]

There is a strange poetry in how the bizarre way of writing the letter “x” (completely missing the essence of the x) is echoed in the bizarre, overly complicated way of solving the problem

[@geirbalderson9697]

How about more confusion as the back to back 'C' kept distracting me! Why not plain old X????

[@Humanity101-zp4sq]

@@geirbalderson9697 x represents 'multiply' in mathematics.

[@mike60521x]

@@Humanity101-zp4sq maybe I am old school, but parenthesis, *, x, or a "dot" was always used to signify multiplication we always used an cursive x to represent variable x - so there was never any confusion I have never seen someone use 2 Cs to represent an variable x until this video

[@geirbalderson9697]

@@Humanity101-zp4sq I understand that. I was agreeing with q-tuber.

[@lefantomer]

If you can't and you still want to go to Harvard, go over to the Admissions Office and plunk down a few thou$and -- much much less than for admission to the college as a full daytime live-in student -- and sign up for Extension School. That's the night school. Pick your concentration if you want a degree. You will attend classes at night run by full professors (unless they've changed the policy), will have to do class work and term papers and a thesis like everyone else, attend Commencement and get a Harvard degree for a fraction of what you would have had to pay had you been "accepted".

[@WilliamFairfield-qq2tq]

Just sate your african American and half native American that studies Islam. Bosh your in full scholarship who's parent is a Jewish convert

[@jbourdette]

The choice of 8 and 4 looks like magic. Be more formal using Gauss theorem about integer roots and soon get -2, the rest is easy, e.g. use Ruffini and Bhaskara.

[@mathis8210]

It looks like magic, because it is. He is just guessing the solution and calling that the "proper method".

[@ericr2646]

I worked out before I saw the youtube, on the right hand side of the screen, first of all X would have to be a negative number, so lets try X = -2, -2 squared is 4, -2 cubed is -8 take 4 - - 8 = 4+8 = 12, of course it is good to watch the youtube in case X is not so obvious.

[@DoubleplusUngoodthinkful]

This is the first time I have ever seen the letter "X" drawn as back-to-back Cs.

[@robertvavra414]

Gives new meaning to 'extasy'. (sorry)

[@ChristianRThomas]

Really? Are you still in your pram?

[@DoubleplusUngoodthinkful]

@@ChristianRThomas "pram"? Are you still in the 1800s?

[@ChristianRThomas]

​@@DoubleplusUngoodthinkful Don't be silly! I'd be calling it a perambulator if this were the 1800s.

[@niravelniflheim1858]

Thanks to those explaining that there are 3 answers to find and various methods to get to the complex ones. -2 was always the easy part.

[@brainhuman7609]

The first thing to remember is when there is a power 3, you need 3 answers, that could be enough bring you to a uni. The rest is what method most students could find them if they remember rule 1

[@robertfitzjohn4755]

I found x = -2 pretty quickly, but as it's a cubic equation it must have three roots. Surely to get into Harvard you have to do better than just say "oh, the other roots are complex, so I won't bother with them"? (Unless they specifically asked for only real roots, of course.) After a bit of work I managed to find the complex roots, and substitute them back into the original equation to prove they were correct. Then again, I did sit the Mathematics entrance exam for Natural Science at Oxford, some decades ago.

[@TonalWorks]

Yeah, that was a wierd assumption indeed.

[@lefterismagkoutas4430]

it's not that hard to find the complex solutions in the state that he left it, hell in most college classes they will leave it like that because it is almost implied.

[@ila9063]

There is no entrance examination for Harvard University.

[@Cotronixco]

Yup. If there were, nobody would attend.

[@Caravanseraii]

The guy just wants to get views duh

[@luciu1001]

Guess the x=-2 Factor the expresion using polynomes rules Solve for cases Way easier

[@tudvalstone]

Writing 12 as a sum of power of 2 just shows you knew the answer. It's easy to eyeball -2 as an answer and make up a "solution". What we need is a way to solve any such equation, not just one where we know the answer from the start.

[@magnadolosfs2707]

That's only one of the answers

[@MandyFlame]

Factor directly as (x+2)(x^2-3x+6)=0 (By inspection x=-2 is a solution, then just work out the other bracket)! Then solve the second bracket using the quadratic formula to find the complex roots.

[@TonalWorks]

Yeah, I did exactly the same just by inspection/ reasoning: (x+2)(x^2 + ? + 6) = x^3 - x^2 + 12. Btw: Why is he only solving for real x's? It's not obvious from how the equation is presented in this video. I hope the Harward problem was more clear regarding the set of solutions.

[@santiagoferrari1973]

Much better than the long way in the video

[@ТочкиЛагранжа]

Even before I started to think, my intuition said me to check negatives. Then, it is a sum of square and third power. 4+8, so, the answer is -2. This is the solution if the problem is supposed for 12-years old children. For Oxford, you also need the complex ones

[@joseluisf7808]

Even for Oxford, nowadays, I have been told that you should approach problem solving from a non-binary gender perspective as well.

[@MrKovalius]

Using theorem about the rational roots of a polynomial with integer coefficients, you can guess that -2 is a root of the polynomial, which is obtained by moving everything to the left side. Then, by Bézout's theorem, you can divide the polynomial by the binomial (x+2), and then you get a quadratic trinomial.

[@kumertd]

Gosh, what does higher mathematics mean in 2024..?

[@cookiemonster-nk3xb]

Do you want a fresh one..mind your manners. If you want hard problems go to Presh Tell Walker at Mind Your Decisions 😂

[@JustinTime-fr8fo]

imaginary math is so diverse and non discriminating it makes the word a better place.

[@The_Mega_AnalistSci]

I just do not know why is he doing that and not doing by Ruffini´s method which is x10 times easier and you get the result faster.

[@lefterismagkoutas4430]

this is actually the quicker method if you know what a^3+b^3 is equal to, without needing to use any other method, he just does it very analytically.

[@The_Mega_AnalistSci]

@@lefterismagkoutas4430 Have you even seen Ruffini’s method.

[@bookert2373]

X² - X³ = X²(1-X) = 12. X² positive, so 1-X must also be positive, ∴ X < 1. For negative values, say X < -M, you have X² > M² and 1-X > 1+M, so 12 = X²(1-X) > M²(1+M). Plugging in values of M of 1, 2, 3, etc immediately shows M must be < 2, and that M = 2 (ie X=-2) is a solution. So reduce degree by dividing original by (X+2) to get quadratic and easy from there. This isn’t ’elegant’, but it only uses elementary algebra knowledge. It’s really all about finding that one solution X= -2 and reducing the poly degree, no matter how you find/guess that X=-2 is a solution,

[@celestine5340]

yeah, that's exactly how I did it.

[@bookert2373]

@@celestine5340 yes, one either guesses that -2 is a root or creates some simple inequalities as we did to find -2 is a root. After that, it’s easy. Knowing that there has to be at least one real root (since degree is odd) is a hint that looking for a real root is probably a good idea.

[@nobodynemoq]

come on, it took me a split second to figure out it's a small negative number and another second to ensure it's -2

[@magnadolosfs2707]

Thats right, that's exactly what I did too but you need to have 2 solutions

[@TheOldeCrowe]

1. kind of easy to see thar x² - x³ = 12 when x = - 2. 2. synthetic division then gives x³ - x² + 12 = (x + 2)(x² - 3x + 6) 3. the quadratic formula gives the two additional complex roots x = ½(3 ± sqrt(9 - 24) = ½(3 ± sqrt(15)i).

[@christianeb8494]

Tout simplement.

[@suityreza84]

You are doing very well . Thanks for doing a good work like this . Mathematics is the queen of science . This kind of problems are very important.

[@panszarlotka]

It can be done more simply. Just decompose the number 12 into factors. It will be 2 * 2 * 3. One of the factors will be the absolute value from X. Then just substitute all the factors into the equation and see if it is true for -2 or for -3.

[@kjetilskotheim1712]

If we assume x is a whole number.

[@jonr3198]

​@@kjetilskotheim1712it's a relatively safe assumption if you're getting an exam with no calculator. That said it takes a few seconds to do this and if you got the answer you're done and if not then do it the complicated way

[@saschaganser9671]

@@jonr3198 You must see that -2 is not the only solution directly. There must be complex solutions too. If you write -2, you didn`t even do half the job.

[@stevenjeng1237]

It is quite simple! One root is -2, so (x+2)(x^2-3x+6)=0, then you can find 3 roots

[@christianeb8494]

Et oui! Concours d'entrée pour Harvard? J'ai fait les Mines d'Albi-Carmaux, c'était tout de même un autre niveau.

[@user-ox4ii2bw6x]

factor by grouping (1-x)x^2 =12 12 is composed of a perfect square, 4 * 3, the function is also a negative degree polynomial so a solution exists on the left side, lets try -2 since its the other value that squares to 4 (1-(-2))*4, innards = 3, done.

[@MartelloClaudio]

Smart solution method. I used anothet method: starting from X^3-X^2+12=0, I used Ruffini's solution method by just seeing that X=-2 is one of the solution of this equation. Then just follow Ruffini rule and get to the same solution in a while

[@TheNizzer]

-2 seems obvious. Then divide x3+x2-12, by x+2… then show that quadratic has no real roots….then find the complex roots. Or sketch the graph with intercepts and max/min and or points of inflexion….to show only one real solution… Seems like a beginner’s question to solving cubics.

[@christianeb8494]

Tout simplement.

[@IITJEEAspirant2025-fj6hj]

x²-x³=12 =>x²(1-x)=12 By trail and error , Factors of 12 are 3×2×2 Now to get , the answer is a positive number But the factors we get x² is positive but there is a problem with (1-x) if 1>x then we also have to consider x in the range of negative number and we should also connect with the negative Factors Hence,we consider the Factors to be -3,-2,-2 By trail and error if we put x=-2 We get 12!

[@ftgsoftware8119]

With the correct answer so obvious by inspection. this reminds me of a multiple choice question on a New York State Science and Engineering Scholarship exam many years ago. You could multiply two long integers (wasting many minutes) or noting that the product ended in 6, just multiply the two rightmost integers in your head.

[@jim2376]

By inspection -2 is the real answer. There are 2 complex answers as we can see by running the quadratic equation on the other factor of x^2 - 3x + 6.

[@mattpanz609]

Me: I want to go to Harvard to be a history major Harvard: here, take this math test

[@stevetaylor5382]

I want to go to harvard to be a physics major. Harvard: Here, you have to take this class on the history of art.

[@hewitc]

Harvard: Take these wire instructions to your Dad. Tell him to solve $1,000 squared times $5.

[@RexxSchneider]

Looking at x^3 - x^2 + 12 = 0, it should be clear that x^3 has to be a negative quantity, since x^3 minus a smaller negative quantity is -12. That should quickly indicate x = -2 is a solution. Then x^3 - x^2 + 12 = (x+2)(x^2 + mx + 6), and we can soon see that m = -3 works (quicker than polynomial division). So the other two solutions come from x^2 - 3x + 6 = 0, which are x = (3 ± i√15)/2. Trivial.

[@DanoshTech]

Bro for Harvard entry solving a cubic equation is very basic and i fell you should be required to find the complex solutions which are (3+-i√15)/2

[@MervynPartin]

I like it when exam questions say to find an answer by some specified method OR BY OTHER MEANS. It makes things so much easier. I once had a half-hour ballistics question in an exam paper to solve by graphical or other means so that opened the door for me to use calculus and finish the question in 5 minutes, giving me more time for the rest of my 3-hour paper (which I finished in 2 hours). I just worked out the answer (-2) to this video's question in my head looking at the thumbnail title, but being long retired, I'm glad I don't have to sit exams any more, not that I wanted to go to Harvard anyway- too long a commute.

[@vict85]

x² - x³ is negative for x > -1, and the function is monotonically decreasing. Therefore, the function has a single negative real root. It is trivial to identify -2 as the real root. The complex root requires a polynomial division. Obviously, there exists a closed formula for the roots of any cubic polynomials, so...

[@nicolasbourbaki8896]

you should realize the answer x =-2 directly just by looking at it

[@RaceSmokie]

Totally agree, it is obvious that x can not be 1 or 2 or 3 ,,, it must be in minus ,,, so -1 does not work, -2 ,,, voilà! What a waste of time with all these calculations, and who cares about complex roots, what is the use of it?

[@wolfrogamer6116]

You see if you don't do this type of work you must realize that what your doing is not math but telling the answer or guessing the answer from your own experience

[@RaceSmokie]

@@wolfrogamer6116 what's the point of writing a dissertation on something totally obvious at first glance. Is it that what they waste time on Harvard, to learn how to blablabla on worthless things or they learn how to make elephant from the fly. Where you will use all of these equations? I dont get it.

[@wolfrogamer6116]

@@RaceSmokie haha so you think the study or math is useless or who study physics or maths are just a bunch of nerds Well in that case you are quit wrong because this "useless equation" are used everywhere. At least every where in the digital world. You see we talk to our computers even our mobile phone through these so called useless equation and if math had not existed how you would have gone through space or even figured out how old something is. I don't know about you but I am a computer science student and I have to create equation for my requirements and I use math to calculate almost everything. And I said that work is nessesary, I said it so because through these steps you can prove that you actually did it. You didn't just cheat off someone else. Now I do get your point that if you just want to know it by your self, you don't want to prove anything to anybody but even then you are still performing these steps and by experience you know the answer.

[@RaceSmokie]

@@wolfrogamer6116 actually you are wrong bcz I dont think that math is useless. now, tell me where you have to use whole that process and what is use of it if you just can see the result without any calculation, what is a point to do all of this. Exactly where will be useful to do this equations. What to prove? Obvious things? Pure waste of time.

[@Cotronixco]

Your X is a reverse C followed by a C, and your 1 is a 7? Why?

[@andytraiger4079]

Inspection reveals X must be a negative number for a real solution... and a very small negative number because X^3 blows up faster than x^2.

[@tensor131]

Factorising gives x^2(1-x)=12, and associating the x^2 with the factor of 4 from 12 immediately yields the integer root. From there the quadratic quotient is easily found yielding the complex roots. Here in the UK that would be an easy question for a16 year old doing further maths A level.

[@bladbimer]

Well honnestly i think this can be helpful to detail on why you deconposed 12 into 8 + 4. It's logical once you did some work before to notice that -2 is a simple solution to find once you factor it x^2(x - 1)=4×(4-1) and so on. I think that what is valuable is more to explain the process more than just apply rules. Like maybe listing things that are important to know such as a^2 - b^b = ... a^3 + b^3 = ... because the value is to see how people are able to see patterns based on the rules they know like those ones.

[@algorithminc.8850]

I look forward to checking out your channel. Subscribed. Thanks. Cheers

[@higher_mathematics]

Thank You Mister. Have a great day!

[@baselinesweb]

Baffled everyone? You only have to substitute for x to get the real root, divide to get the quadratic and then use the quadratic equation to get the complex roots. Don't oversell.

[@mrironmaiden100]

After finding x=-2, it's simpler to divide the polinomial for (x+2) using Ruffini's rule

[@surfer_guy942]

Use the rational zeros theorem and get x = -2, x = (3 +- sqrt(15) i ) / 2.

[@devastationofmankind3495]

So...for everyone thats not understinding the joke...yeah no me neither i quite literally have a grand total of ZERO idea what he's doing after the 4 step...but hey, neither do you...

[@stevenjohnson1143]

Easy -2 squared is 4 minus -8 when subtracting a negative number flip both signs gets 12

[@bobs1224]

This can be simplied. We can factor x to get x^2(1-x)=12. Since 4 x 3 = 12, we can write x^2=4 and 1-x = 3. Solving for both gets us -2.

[@jaycathedral9079]

Trig in high school and secured transactions in law school had one thing in common: Over the ensuing decades, I perhaps used each discipline twice. Algebra and calculus were a little more practical and I have yet to need to know the maximum area of a circle that can be inscribed in a square with a radius of X. Then again no one has asked me about the gram molecular mass of a molecular substance either. I took a lot of math and science and am glad I did. Not sure why. . .

[@mbogucki1]

Have you been asked anything can't be figured out by Google or ChatGPT?

[@mike60521x]

Math trains the mind to think efficiently as well as expands the thought process to use different operations to solve complex equations - while you may not need specific instances of it in your present day life, the knowledge has structured your brain to think in an orderly process and question all parameters of a problem

[@loicgeeraerts]

I have another approach that does not require knowing the formula a^3 - b^3 and which allows you to find the solution -2 almost directly : x^2 - x^3 = x^2 (1 - x) and 12 = 2^2 x 3 but the problem is that we choose x = 2, we don't have 1-2 = 3. No problem, just replace the 2 in -2 because 2^2 = (-2)^2. To recap, x^2 (1 - x) = (-2)^2 (1 - (-2)) , so -2 is a solution. Then, just factor by x - (-2 ) to get x^2 - 3x + 6 and check that there are no other solutions.

[@yatharthmcoc9752]

X = -2 is a root, hence X+2 = 0 must be a factor of the expression Divide the whole the expression by X+2, the 2 degree equation that we get has 2 of the roots, solve that and you get the remaining 2 roots That’s how my teacher taught us

[@hippophile]

x = -2 is a root after a little thought, and then x^3 -x^2+12 must have (x+2) as a factor soit's easy to find the other quadratic factorby looking at th coefficients, and thence using the formula you can figure out the complex roots.

[@bartermens8219]

You also draw 2 graphs x² ans -x³. And see where they are 12 apart. Sounds much easier than all the hokus pokus with x.

[@Hyperion1722]

Easy. Place X squared on the other side and you have a quadratic formula (X squared -12). The quadratic formula is ax squared +bx + c.

[@jerrypaquette5470]

It is pretty obvious that one solution is -2. Therefore X+ 2 must be a factor of X^2 - X^3 -12. Multiply -1 you get X^3 -%^2 + 12. Using division (X^3 _X^2 +12)/ (X+ 2) you get X^2 - #X + 6 X^2 - 3X + 6 = 0

[@paulglover6862]

Thanks for the video. I worked it out in a few seconds.

[@ethannguyen2754]

Rational roots theorem gives you x = -2 Divide the expression by x+2 to get x^2 - 3x + 6 and solve for the remaining roots with the quadratic formula. This is how most people who could solve this problem would do it. Your solution is interesting, but looking for those kinds of solutions would probably waste a lot of time on the exam. Especially if the more basic strategies for finding roots of polynomials work well enough.

[@Stafford674]

x^2 - x^3 =12 X^2(1-x) =12 So x^2 is a factor of 12 and 1-x is a factor of 12. The following are factors of 12: 2,6,-2,-6,4 -4, 3, -3. X^2 must be positive, and a square. x^2 must therefore be 4. The roots of 4 are 2 and -2. If x was 2, 1-x would be -1. If x = -2 then 1-x = 3. x^2(1-x) = 4 x3 = 12. So x = -2. Simples.

[@BruceLee-io9by]

It's easier to break down the polynomial using Ruffini's method.

[@alexlucky3338]

Exactly my thoughts) but it's good to have an alternative method in mind as many people are "allergic" to polynomial division 😅

[@EddyGilley]

I liked Raffini in #LionKing. When he held up Simba on Pride Rock before all the Pride Landers to see, it gave me chills. Circle of Life. #RaffinisRule 😉 Semper Fi

[@wilson2455]

certainly took the extra long way for something I did (in my head) in about 10 seconds !!

[@marcozoboli]

You can easily solve that equation noticing that x=-2 is a solution and then factorizing using the Ruffini's rule.

[@danweingarten7674]

Far easier approach, factor both sides: (1) LHS: x²-x³ = x²•(1-x) (2) RHS: 12= 4•3 = 2²•3 From (1) & (2) we have x²=2² and (x-1)=3. Solving the second equation we get x=-2.

[@tommymorton4939]

1) X = -2. 2) Divide to (x+2), the result is square eqv. 3) Find two more complex roots.

[@geofwassell]

Simple mental arithmatic gives an answer of minus 2. I doubt that Harvard is this easy. Oh the complex issue: the problem itself ridicules the reason for complex numbers.

[@robertvavra414]

Enjoyed the video. I haven't studied math in many years. I eyeballed the '-2' answer in a few seconds, but assumed there was another answer.

[@wisemang73]

Passes harvard math exam. Fails 1st grade writing

[@gabbleratchet1890]

It's not really the entrance exam for Harvard. There is no such thing. They look at your grades and SAT/ACT scores.

[@NainoLoL]

​@@gabbleratchet1890 Or alternatively, your wallet 💰

[@jayrussell3796]

​@@mxm5783I actually commented that 2 days after your post...that scribble drove me crazy😅😅

[@robertveith6383]

Original poster, you failed First Grade Writing by not writing sentences.

[@gaynzz6841]

@@gabbleratchet1890 No, they look at your bank statements.

[@udhessi3444]

7.27 minutes for an incomplete solution ! It should go like this : -2 is an obvious solution, so divide x3-x2+12 by x+2, that's x2-3x+6, which you simply solve with the quadratic formula, hence two more roots [3+i.sqrt(15)]/2 and [3-i.sqrt(15)]/2.

[@AlexEckelberry]

Just do some guesses. Guess by pretending x = 2, then x = 1, then -2, etc. to find it. It doesn't have to be so complicated. And you can also just go to Harvard extension online these days - there's a 4 year degree online that if you pass the first class, you're in. It's hard but it will be a Harvard degree.

[@jbjohnson4728]

Why can't you do x^2 minus x^3 equals x^-1 = 12. In basic alegbra (my level), I was taught that x^-1 = 1/x (1 over x). Set that equal to 12 and solve for x and you get 1/12. If I plug in 1/12 into the original equation I obviously don't get the right answer, but it seems like my method should be right based on the basic algebra I was taught.

[@DG-sy3rv]

It is a 5 minutes exercise. x2=12/(1-x) >0. So, 1-x>0 and sqrt[12/(1-x)] is an integer number. The result is obvious.

[@DavyanHatch]

Lots of people don’t realize this is satire…

[@t1mmmmo]

It seems to me that the "flash of inspiration" needed to reanalyse 12 as 8 + 4 is harder than seeing by inspection that x=-2 and then substituting that in. Or, to put it another way, if you didn't know the "trick" in advance, why would you think of reanalysing 12 as 8 +4 to progress as suggested?

[@warrenbrown8670]

Did it in my head in under 10 seconds after I saw it. Interesting to see the process but some things you just see the answer.

[@MariusBezuidenhout31337]

Order of operations (exponents before multiplication) would mean that -2 squared = -4. If x = (-2) it would be correct otherwise it would be x ≈ 2.68

[@CuriousFocker]

I was going to post exactly the same thing. It's amazing the number of KZbin maths channels that get things wrong.

[@Gary.Holmes]

This is a really interesting point. I hadn't considered where the minus sign belongs in -2^2 but I do see the ambiguity, that -2^2 is not (-2)^2 . Is that what you mean? Thanks for making my brain itch! 😀

[@FEPractice123]

by convertin x*x(1-x)=12 you will know (by inspection) that the solution must be a number less than 1. Thus, using x=-2 you have the solution in less than 30 seconds.

[@BRAYANEMPEROR]

X² factor and test the numbers that their multipy gives 12 then its 4 for x² and 3 for x-1 and also negetive numbers are not alow becuse power 2.

[@Noviscentur]

-2, элементарно. Даже не решал, сразу подставил ответ, он на поверхности.

[@andredcavalcante]

That's is a simple equation: what number the square is greater than the cubic form? Well, this number must be negative. Well... trying -1, square(-1) minus cubic(-1), 1 + 1 = 2, don't equals 12; trying -2, square(-2) equals 4 (promissing), cubic(-2)=-8, 4+8=12. I got it! The first real root is -2. I know may there are 2 other real ou complex roots, but I'm ok with these rapid and simple thoughts.

[@jofel131]

Correct but very long. Surely the solution must be negative, because x^n is strictly increasing as a function of the positive integer n > 0, if x > 0. Morover the solution must be small, because x^2 - x^3 increases very fast as x < 0 decreases. So, before calculation, just try with -1 and -2. Alternatively you can write x^2 - x^3 = 12 as x^2(1-x) = 4 * 3 and realise that 4 = (-2)^2 and 3 = 1-(-2), using intuition. Since the function x^2 - x^3 strictly increases as x < 0 decreases, it can be equal to 12 at just one real x, so the other two solutions must be not real.

[@superserioes]

I was very bad at math at a normal school. Of course I wouldn’t have passed this test. But I’m positively surprised that not only did I guess -2 fairly quickly, but I also completely understood the proper solution here. For me that’s something, considering this is effing Harvard.

[@gaynzz6841]

You still wouldn't go to Harvard.

[@superserioes]

@@gaynzz6841 You don’t say

[@mattlm64]

I did it via an educated guess that it was going to be -2. I guess it's one of those annoying "show your workings" questions.

[@frenzy1111]

There is in fact no question just an equation but assuming that the question is to solve for X, then X should be defined as an element of a number set. X€R, N or C or similar. Without this information we can't hope to answer any implied question.

[@zahidabdurrahman627]

1:55 if you can't give the method to work this point out mathematically your entire method is pointless. You could just deduce the answer without the long windedness of the working out, instead of going through all this while you deduce this part with no mathematical formulae of working it out, if the numbers in the question weren't so whole or rational your method would be useless.

[@moho_sama]

I don't know the american standard about mathematics but even a simple student from any highschool in italy can do this. Just find something make zero so x=-2 and just go on with Ruffini's technique. And is the same as the video x1=-2 x2,3= complex. And at university they teach me to converting the negative root to i(sqrt) of 15 so x2,3 will be: 3+i(sqrt)15/2 and 3-i(sqrt)15/2 Very impresses cause i think is simple but i repeater i'm in italy and i don't know about the mathematical standards in us.

[@qzwxecrv0192837465]

wait, many people failed this simple math problem? i did it in 10 seconds in my head: (-2)^2-(-2^3)=12. Given you have to subtract the second number, the root number has to be negative in order to ADD it to the first number. so simply, using 1 for x, we get 1-1=12 WRONG, with two we get 4-8=12 WRONG but change the sign to + would make it work, thus -2. no need for long formulas, quadratic equation, etc. for such a simple math problem.

[@darkboi5422]

For JEE it is taught that in 99% cases A root for a cubic equation is 0 1 2 -1 -2 and 0.5

[@alankwellsmsmba]

I get it, but in my experience (I'm an MS/MBA from a mid-tier Catholic university), vetting candidates who can successfully solve the equation doesn't seem to help Harvard find the best and brightest based on the quality of output, especially in their school of law.

[@АлёшаИнкогнитов]

x² - x³=12 1-x=12/x² x=1-12/x², so X is rational. Also X

[@prathammittal3870]

I solved this mentally. Since result is +ve than x2 is greater. So either 0

Yeah, the real root is quite elementary. Figuring it out took me just a couple of basic thoughts. The complex ones, weeell…

[@ShifatulIslam-cy3ne]

x^2-x^3=12 the answer is x^2(1-x)=12. x^2=12\1-x . x=√12/1-x =2√3. Thanks 😊😊😊

[@ShifatulIslam-cy3ne]

❤

[@johncusson5703]

Looking at the equation, someone may surmise that the answer must be a small number. A small positive number won't do, -1 does not work, -2 is the answer.

[@lloydnicholls1439]

Sweet! I figured out -2 in a minute without watching the video! Haven't a clue what all that maths is on this video, though!

[@pmbpmb5416]

At a second glance -2 , if asked to show workings they would roll around the floor .

[@satautenyo]

then the solution depends on FIGURING 12=8+4... in this case is also valid FIGURING that x=-2 is the first solution (trying integers by hand) and dividing the cubic equation by (x+2) which yields x^2+3x+6=0 and then the roots are 3/2 +- sqrt(15) i /2 as said by other mates... I don't think mine nor author's are valid and FORMAL solutions anyway...