Entrance examination and Olympiad Question in 2022. 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
As apparently some others, I (with a phd in maths, but out of science for years) am a bit confused about the solution. If you guessed t=5, to prove it you can just enter it into the formula, no need to factorise. To factorise, just do polynomial division, no need to guess something wildly like the separation of t^2 into -5t^2+6t^2 etc. Polynomial division tells you algorithmically how to do it. Yes, there do exist theorems (and this is well beyond ordinary school maths) that if such a polynomial (with integer coefficients) has a rational (a fraction) solution it must already be an integer and then it will be a divisor of the constant term (155). Hence, it is a good strategy to try divisors of 155 (aka: +/- 1,5,31,155). If your teacher likes you, one of them will work. Once you factorise to degree 2, the other solutions can be found with pq formula or quadratic extension, however you name it. BTW, there do exist (very complicated) methods for degree 3 and 4 too. BUT: There is no indication that 2^x is actually rational. To state it clearly: this was just guessing and hoping for the best. Of course, if you have no better idea, it is worth a try.
I kinda disgaree with "beyond ordinary school" Here in Indonesia we learn about rational root theorem in high school. I believe Cambridge Further Math A Level or IB DP also teach about rational root theorem
@@thejelambar82 I sadly believe you. Of course I was talking about Europe and Germany, maybe US, where I strongly doubt you learn this before joining university. Comparing to Cambridge, well... Just lets say this is not what 99.999% of pupils experience. It is a well known fact here, that the maths education in Asia is much, much better than it is here. Personally, I'm a bit curious if there are other topics which are teached better in the western world than in Asia in return, or.. if our education is generally worse. Again, I fear I know the answer.
Start with the equation 2^X = Y. Given: Y + Y^2 + Y^3 = 155 This can be rewritten as: Y * (Y^2 + Y + 1) = 155 Find the factors of 155: 155 = 155 * 1 155 = 31 * 5 Test the possible values of Y: Y = 1 does not work. Try Y = 5: LHS = 5 * (25 + 5 + 1) = 5 * 31 = 155 Since the left-hand side matches the right-hand side, Y = 5 is correct. Thus, we have: 2^X = 5 Solve for X: X = log(5) / log(2) So, the value of X is log(5) / log(2).
That's the way I did it. The key is realizing that 155 is the product of two primes (5 and 31), which simplifies matters considerably. I hate it when presenters introduce terms (add this, subtract this, etc) without explaining why they chose the particular values that they did.
@@silverhammer7779 Dude the thing he did is simple grade 9 mathematics. He simply did factorization of a cubic equation, any person who has graduated high school should be able to do that. I don't think it needs explaining
😮Independently followed your methods through the cubic equation, but then quickly realized t must equal 5 because only multiples of 5 add up to 155. Synthetic division could have been employed, but was unnecessary. Then solved with logs as you did.
Ah, yea....let's just guess 4, no 6...no let's go with 5. This could be a Dilbert solution....where the next to last step is: "Then a miracle happens " As stated ....there is NO Harvard entrance math problem
Can we do it like this for faster answer but the ans would be near the original ans not exact 2^x + 2^2x + 2^3x = 155 2^6x = 2^7.3( approx ) On comparing, 6x = 7.3 x = 1.21 ( this is my ans )
Math 55 covers essentially an entire undergraduate curriculum. And people have been known to place out of it. So there would still be plenty of things to keep you entertained for the next few years.
Clearly not good Need to eliminate fake intelligence. Teach life skills to high schools. Complete teachings Buying insurance. Car home life insurance And different types Something actually helpful?!!!!??
Take the step from t^3+t^2+t=155 to t^3-5t^2+6t^2 etc. This seems like a totally random step. Why choose -5 and +6? I love to know the reasoning behind this step.
By the method of staring at the equation he noticed that 5 is one of the roots. He then chose coefficients so that he could factor out (t - 5) to check if there are other real roots. You can as well group like this: (t - 5) + (t^2 - 25) + (t^3 - 125) = 0 (t - 5)(1 + (t+5) + (t^2+5t+25) ) = 0
@@georgerodionov5941 I have another proposal: after guessing that 5 is the root, we can notice that t, t^2 and t^3 are all monotone increasing functions (at least in positive t, which is what we are interested in), so their sum is increasing too. And then there cannot be more than one intersection with the horizontal line at height 155.
My brain: There’s a 5 in here, right? Me: Maybe. But you’re 40 years old, you might have epilepsy and you had an actual seizure 4 weeks ago. My brain: So that’s why I understand absolutely nothing else 😅
Yes, and solve the quadratic equation in 2^x as complex conjugates. Then take the log base 2 of them to yield an infinite number of solutions related to 2 main values.
I think that you are making these up as you go along. Is this realy from Harvard, was the other one realy from Cambridge? They are good fun but perhaps not as hard as your titles suggest?
You use a solution to a cubic equation which is quite difficult after forming the equation t^3 + t^2 + t -155 = 0. On inspection I would set t to 5 and solve. 125 + 25 + 5 - 155 = 0.
Once yo make the change of variables you can get the real 5 solution by examination. You do the remaining part to show that the other solutions are complex and that you have not overlooked some not-so-obvious real solution
I fully agree with @fibonacci_fn ! Since the equation is degree 3, you have to "guess" an integer solution for Y, and Y = 5 comes rapidly (especially when you notice that 155 = 31 x 5) ! My remark is : we have to show that 5 is the only real solution (and then ln 5 / ln 2 is the only x)... For that purpose, we see that f (x) = 2^x + 4^x + 8^x is strictly increasing on R, and the TVI says that it takes a given value 155 for at most a single x. (By the way, this avoids the calculation of (Y + Y^2 + Y^3 - 155) / (Y - 5), and the "∆ < 0" of the quotient...). Thanks for your videos !
Here in India we do this kind of problem even harder in 7th grade even in remote villages schools 😂😂 "VANISHING METHOD " ASSUMING AND PUTTING, X=5😂😂😂😂😂😂😂😂😂😂😂
Lets not throw out the complex babies w/ the bath water! So, complex roots are given by t²+6t+31=0 ⇒ t=√31·[(-3±i√22)/√31]. To evaluate log of t, we choose the branch cut of the log to be the non-positive real half-line & restrict the arguments of the complex numbers to lie between -π & π. Then, log₂(t)=log₂(√31·[(-3±i√22)/√31])=log₂(√31·e^(±iθ)) for θ=cos⁻¹(-3/√31)≅.681π ⇒ log₂(t)=log₂(31)/2±iθ/ln(2) where "ln" is the natural log.
@@oahuhawaii2141 You can generalize the solution I mentioned by adding 2nπ, for n any integer, to the expression for θ, but you have to consider consistency & functions becoming multi-valued. For one, the real solution mentioned in the video, should then be generalized to t=5 → t=5e^(2ikπ) ⇒ x=log₂5 → x=log₂5+2ikπ/ln(2) for k∈ℤ & you should choose k=n, so that all 3 solutions are consistently in the same interval. And, why not generalize further t=2^x → t=[2e^(2iπm)]^x for m∈ℤ ⇒ x=ln(t)/(ln2+2iπm) So, to at least be consistent w/ the real solution presented in the video, I chose the fundamental interval for the complex plane.
Your math work isn't rigid. You should be writing that 2^x = 5 is one root to the cubic form of (2^x)^3 + (2^x)^2 + (2^x) = 155 . You still have to find x, and see if the other 2 solutions to the cubic equation are real or complex conjugates.
Si poteva risolvere l'equazione cubica anche scomponendo il polinomio di terzo grado con il metodo di Ruffini, notando che tra i divisori del termine noto, 155, vi è proprio il valore t=5
Let 2^x= y. So y+y^2+y^3=155 So y(y^2+y+1)=5*31 After close observation,Equating y=5, gives (y^2+y+1)=31. So conclusion y=5 =2^x x = ln 5/ln 2. Very easy to get real solution 😂
how about this? 10 print "higher mathematics-can you pass harvard college entrance exam?" 20 sw=.1:x=sw:goto 50 30 dgu1=(2^x+4^x+8^x)/8^x:dgu2=155/8^x:dg=dgu1-dgu2:return 50 gosub 30 60 dg1=dg:x1=x:x=x+sw:x2=x:gosub 30:if dg1*dg>0 then 60 70 x=(x1+x2)/2:gosub 30:if dg1*dg>0 then x1=x else x2=x 80 if abs(dg)>1E-10 then 70 print x higher mathematics-can you pass harvard college entrance exam? 2.32192809 > run in bbc basic sdl and hit ctrl tab to copy from the results window
