Japanese | Can you solve this ? | A Nice Math Olympiad Algebra Problem

4 місяці тому

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КОМЕНТАРІ: 39

@user-tl9bq7gd9v День тому

Lovely work!

@ms9035 4 місяці тому

After reaching the following equation, it is better to do the rest of the operations in the following simple and easy way to avoid the complexity of the answer. ( a+b)^2 + 2( a+b) = 8 (a+ b )^2 +2 (a+b) + 1 = 8+1 = 9 ( ( a+b) +1 )^2 = 9 (a+b)+1 = 3 or ( a+b)+1 = -3 a+b-2 =0 or a+b+4 = 0 And then we will continue to write the answer in the same way as you have done. Thank you my best friend

@dmitrynikiforov8198 4 місяці тому

Absolutely, so did I too.

@michelnkoghe7363 Місяць тому

After obtaining équations 1 and 2 as following Eq1: a^2 + a + b + ab = 3 Eq2: b^2 + a + b + ab = 5, Why not substrat them directly by doing Eq1 - Eq2? Thus a^2 - b^2 = 3 - 5 = - 2; (Eq3); (a + b)(a - b) = -2 = (-1).2 = (-2).1; the 1st couple ((a+b),(a+b)) brings from Eq3: a + b = -1 (Eq4) and a - b = 2 (Eq5); by adding them, we get 2a = 1; a = 1/2; wuth Eq5: 1/2 -b = 2;; b = 1/2 - 2 = -3/2 => 1st couple (a,b) = (1/2, -3/2); thé 2nd couple ((a+b),(a-b)) brings from Eq3: a + b = -2 (Eq6) and a - b = 1 (Eq7); by adding them, we get 2a = -1; a = -1/2; wuth Eq6: -1/2 + b = -2;; b = 1/2 - 2 = -3/2 => 2st couple (a,b) = (-1/2, -3/2); remembering that a = √(X - 1) and b = √(X + 1), Squaring both a and b whatever thé couples WE got brings this: X - 1 = 1/4 X + 1 = 9/4; Adding them brings 2X = 10/4 = 5/2; thus X= 5/4 >= 1. Thé solution IS thé same: X = 5/4

@sharatchandrasekhar2711 3 місяці тому

Just substitute y=sqrt(x-1) and collect terms into the form sqrt(y^2 + 2) = (4 - y^2)/(1+y) - 1 Now square both sides and rearrange to get (y^2 + 2)(1 + y)^2 = (3 - y - y^2)^2 Expand and note that what appears to be a quartic actually degenerates to a quadratic with only one non-spurious root. The answer x=5/4 follows.

@gerhardb1227 3 місяці тому

Thank you for your solution: Here another approach: substitute x = a^2 + 1 will result in a quadratic equation => 8a² +10a-7 = 0 a1 = 1/2; a2 = -7/4 a1 will be a valid solution = 1/2

@rosariobravo9165 3 місяці тому

@gerhardb1227 Buenos días. Muy interesante su enfoque. Entiendo que sólo hace una sustitución de variable. Lo voy a intentar. Gracias.

@rezanader5770 10 днів тому

Hi, at the beginning of solution the obtained domain is incomplete. You should also solve 4-x>=0. So the domain is 1=

@jarikosonen4079 3 місяці тому

At 4:01 if you add one on both sides, what will happen? (Numbers on right side getting bigger?) 17:46 Eq. 3, could be solved directly also backsubstituting both a and b. Key seems right substitution.

@MartinPerez-oz1nk 4 місяці тому

THANKS PROFESOR !!!, VERY INTERESTING !!!!

@learncommunolizer 4 місяці тому

You are welcome! Thank you very much!!

@user-xc5os4ep3n 19 днів тому

Икс в квадрате минус один-это же (х-1)(х+1)😊

@knotwilg3596 29 днів тому

Observe that the square of the inner terms equals 2 times the outer terms: (v(x-1)+v(x+1))² = 2x + 2v(x²-1) Set t = v(x-1)+v(x+1); (A) we get t + t²/2 = 4 or t²+2t-8=0 This quadratic equation has solutions t= -1 +/- 3 = 2 or -4. Since t is the sum of sqrts, it's positive and only t=2 is valid. Set u = v(x-1) then v(x+1) = v(u²+2) (A): 2 = u + v(u²+2) or 2-u = v(u²+2) square that, so (2-u)² = u²+2 so u²-4u+4 = u²+2 so -4u = -2 so u = 1/2 So x = 5/4

@rober3072 3 місяці тому

There is a mistake a² >= 0, then a >= 0 or a= 2 then b>= sqrt(2) or b= 1, then a >= 0 and b >= sqrt(2). The solution is OK, but the reasoning is incorrect

@user-pv7jv3dc9s 3 місяці тому

To be shortly solving, let's ander root x-1 + anger root x+1 =t, Than will be= t power 2+ 2time t+=8, » t+1 powe2 =9» t+1 =3 t=2 and etc.

@MrUtubePete Місяць тому

He sure took a loooong way to get there

@Pierre1O 4 місяці тому

Nice solution! Note: Sqrt of any value is always positive, so a+b is always positive. No need for all these inequalities to prove a+b /= -4.

@rosariobravo9165 3 місяці тому

Precioso ejercicio.

@learncommunolizer 3 місяці тому

Thank you very much 👍👍

@yardenvokerol4253 3 місяці тому

It's a very slippery slope to victory. if you take a wrong turn, you will find yourself in a dead end

@gruba4630 2 місяці тому

Condition that square roots must be real numbers is not mentioned in the original problem, you introduced it yourself, so your solution is partial. Also, you spend too much time explaining what should be obvious for this math level. It would be nice to know which Japanese math olympiad it was, year and level. Otherwise, very nice presentation.

@captainteach007 2 місяці тому

by the way, perfect square is not the same as difference of squares

@ronaldnoll3247 4 місяці тому

The result is correct... x=1.25

@nagarajahshiremagalore226 23 дні тому

Pl write what you are telling

@user-im1hv7oc4w Місяць тому

Mi fa male la testa seguirti.😢😢😢😢

@Se-La-Vi 3 місяці тому

Много лишних шагов с преобразованиями

@Nehezra2023 4 місяці тому

In your solution, there is an error in solving (x+1)(×-1)>=0. You forgot (x+1)=0 AND B>=0) OR (A

@comdo777 4 місяці тому

asnwer=3 isit

@user-it6fh7hy6t 4 місяці тому

Профессор! Нормальный учитель распишет такое уравнение за 2 минуты,даже отвлекаясь на чай.

@dmitrynikiforov8198 4 місяці тому

Are you serious ? This is OLYMPIAD problem ? It has taken from me about 2 minutes to solve it. This is the problem for a university matriculant.

@user-it6fh7hy6t 4 місяці тому

Какой же университет закончил наш профессор, если на такое примитивное уравнение он потратил 25 (!) минут?

@dmitrynikiforov8198 4 місяці тому

@@user-it6fh7hy6t Сам удивляюсь :)

@l.w.paradis2108 4 місяці тому

Well, so did I, but these are screening problems. People who don't solve a whole bunch of these really fast do not qualify.

@l.w.paradis2108 4 місяці тому

All of these are screening problems from timed tests. The real problems are multi-step projects involving informal proofs.

@Billts 4 місяці тому

This is penise exercise. I don't understand

@peterotto712 4 місяці тому

Much too complicated - square the original equation twice

@schlingel0017 4 місяці тому

It is more complicated like that and even the. you will still have some square roots left in the equation.

@bdfu4321 3 місяці тому

无聊

@johnlee6304 4 місяці тому

No need to write down that much