Can You Pass Harvard's Entrance Exam?
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- เผยแพร่เมื่อ 26 พ.ค. 2024
- Entrance examination. If you're reading this ❤️.
What do you think about this problem?
Hello My Friend ! Welcome to my channel. I really appreciate it!
@higher_mathematics
#maths #math
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For the guys who directly substituted x= -2🤙
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!"
Yes, but youcan say, that according to Vieta's formulas if there is a rational root, 12 can be divided to it.
In my head in -2^3 seconds.
5 seconds for me. And I have studied history, not mathematic...
But that's not the only solution
The correct answer is: My parent’s just donated $5M to the school.
😅
You have to pass the school as well. Donation can only get you admission. @@aliahmed800
Yup
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
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
👍
That's only one of the answers=FAIL.
but it is mathematically incorrect. you miss 2 of the three solutions...
Solution only works for integers, not real numbers
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.
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
Seeing -2 only took about 2 seconds. For Harvard, you should need to find the complex solution, as well.
Not sure about that.
Wolfram came up with x=-2 under "real" solutions. I agree with @potrahsel4195 .
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
Then it should be a harder problem. I'm dumb. It took me 5 seconds.
@@chrisclub3185 by x+2, not by x+1 ))
I look forward to checking out your channel. Subscribed. Thanks. Cheers
Thank You Mister. Have a great day!
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.
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.
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.
Yeah, that was a wierd assumption indeed.
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.
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.
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.
Exactement.
imaginary math is so diverse and non discriminating it makes the word a better place.
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.
I loved it. brought back my school memories
There is no entrance examination for Harvard University.
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.
Me: I want to go to Harvard to be a history major
Harvard: here, take this math test
-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.
Tout simplement.
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
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
I actually tried a fractional answer at first, but then I remembered that odd powers change signs. Can't say as I'd have figured a complex answer except that Wolfram Alpha listed one when I checked my solution. Neat.
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. . .
Have you been asked anything can't be figured out by Google or ChatGPT?
Use the rational zeros theorem and get x = -2, x = (3 +- sqrt(15) i ) / 2.
It’s intuitive that x = -2, but if x would have been anything else I would have been lost on how to solve it, even though it looks so simple. Thank you for the tour of the solution, I need a lot of practice to become comfortable with those steps.
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
You can factor out one x set it equal to zero and solve it as a quadratic
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
I was going to post exactly the same thing. It's amazing the number of TH-cam maths channels that get things wrong.
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! 😀
Easy -2 squared is 4 minus -8 when subtracting a negative number flip both signs gets 12
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.
you should realize the answer x =-2 directly just by looking at it
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?
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
@@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.
@@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.
@@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.
certainly took the extra long way for something I did (in my head) in about 10 seconds !!
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.
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.
Much better than the long way in the video
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...
It is quite simple! One root is -2, so (x+2)(x^2-3x+6)=0, then you can find 3 roots
Et oui! Concours d'entrée pour Harvard? J'ai fait les Mines d'Albi-Carmaux, c'était tout de même un autre niveau.
Was following you right up to the cubic formula. Not sure how you concluded that.
-2, элементарно. Даже не решал, сразу подставил ответ, он на поверхности.
Gosh, what does higher mathematics mean in 2024..?
Do you want a fresh one..mind your manners. If you want hard problems go to Presh Tell Walker at Mind Your Decisions 😂
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,
yeah, that's exactly how I did it.
@@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.
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.
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...
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).
Tout simplement.
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.
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
You can easily solve that equation noticing that x=-2 is a solution and then factorizing using the Ruffini's rule.
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.
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.
@@lefterismagkoutas4430 Have you even seen Ruffini’s method.
I think this video should be played backwards - it keeps getting more complicated instead of simpler! I solved this in my head in about 20 seconds!
Bro u can also use the hit n trial method
It will be much easier and quicker way to solve this
But i appreciate u for this method also ❤
What? This is just a simple cubic equation
Passes harvard math exam. Fails 1st grade writing
It's not really the entrance exam for Harvard. There is no such thing. They look at your grades and SAT/ACT scores.
@@gabbleratchet1890 Or alternatively, your wallet 💰
@@mxm5783I actually commented that 2 days after your post...that scribble drove me crazy😅😅
Original poster, you failed First Grade Writing by not writing sentences.
@@gabbleratchet1890 No, they look at your bank statements.
well done, cool proof -
Yeah, the real root is quite elementary. Figuring it out took me just a couple of basic thoughts. The complex ones, weeell…
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.
If we assume x is a whole number.
@@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
@@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.
At a second glance -2 , if asked to show workings they would roll around the floor .
x³-x² = 12 ---- x²(1-x) = 12 ---- x
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.
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
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
My thought process, immediately realised that to get a positive number of 12 then X must be a negative. Then guessed -2 and it worked. I can't do all that long maths but in my head it took 5 seconds to solve 😂
pro tip: harvard doesn't have an entrance exam
lol
Haven't done maths since school in 80s, but knew it must be a negative number otherwise a cubed number tsken from a squared number could never equal a positive answer!
And i remembered that subtracting a negative nunber is just adding that number.
So I just tried some whole numbers, first -1, then -2 and bingo!
Who needs to use complicated formula or procedures (I haven't watched the video yet lol).
A hard question would be if instead of 12 it was say 13 or 12.5 haha...
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.
Guess the x=-2
Factor the expresion using polynomes rules
Solve for cases
Way easier
I am very sorry for those that applied Cardano's formula
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.
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".
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.
Do a readable, well-written, straight X ... how hard can it be??
x=(-2)
It’s like a -4 + 16 kind of deal.
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.
I got lost at the formula in the box. Can someone tell us the practical value of this knowledge?
I solved this mentally.
Since result is +ve than x2 is greater.
So either 0
I can't tell you how many times this replayed 😂. The first time, I was looking at the wall for bullet holes
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
After seeing problems and solutions like this, I don't even know how I got through algebra. I must have been shown a shorter way. I don't recall having to do problems this long.
Yeah was cool. Cuz Mom taught me stuff like this in fourth grade. Best Mom ever! 😊❤
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.
I remember doing this exact problem in my college algebra class. I didn't know I could get into Harvard with it LOL.
This is called - Math zigzagging 😆
x² - x³=12
1-x=12/x²
x=1-12/x², so X is rational.
Also X
Hey I'm in! Guessed -2. Job done.
Congratulations to Harvey's University. Home of the Big Harvey's Burger in Canada!
So I looked at this and knew it was 2 or -2. I guess I guess i learnt something in engineering.
My question is, where in life di we use this? I spent 4 yeara of my life learnings matha like this, then the last 10 years beings an engineer, and I've never used any thing I learnt in my degrees except basic trig, basic excel, basic modelling and manufacturing principles aka tollerances and material weights / strengths.
Im seriously curious, how does this question factor into everyone's besides specialists lives?
Why every time he said "you can simply write it like this", it got more complicated
Those who just used the cubic equation and were done with it 🧠 🗿
How can I solve a problem like this when I can't even balance my check book?
-2 est racine évidente, il n'y a plus qu'à faire une division polynomiale par (x+2) pour ramener cette situation à un polynôme du 2nd degré? Concours d'entrée pour Harvard? Pas besoin de monter une usine à gaz.
In china this is middle school maths and if you can't do it, your math teacher will call your parent
Solved it in my head. Super simple. Had to be negative number. -2 solved to 4 - -8 = 12. Sure there’s a sophisticated way but why when it’s easy
I love Harvard, but the MIT in me found the answer in a quarter of the time.
Harvard doesn’t have a separate entrance exam, though… you might find this on an AP test, I guess?
I heard that today is enough to shout “ from the river to the see” and you are admitted.
I mean, I did it my head in about 4 seconds. I know there's a complicated way to do it but doesn't just getting to the answer count?
X^2(1-x) = 12 . The only multiplier of 12 that is a square is 4.
1) X = -2.
2) Divide to (x+2), the result is square eqv.
3) Find two more complex roots.
Well, x^2(1-x) on one side and (-2)^2 - (-2)^3 = 4- (-8) = 12. So x=-2 , it has to be negative and you know 4+8 =12
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.
If square is greater than the cube then X is negative. Started with -2 for X and it worked. Next!!
The real question here is what the uploader was high on that they didn’t get -2 right away by just putting it into the equation …
I worked it out in my head in less than five seconds.
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.
You still wouldn't go to Harvard.
@@gaynzz6841 You don’t say