Japanese | Can you solve this ? | A Nice Math Olympiad Algebra Problem
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- čas přidán 28. 12. 2023
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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
Absolutely, so did I too.
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
Nice problem. Good solution
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.
Lovely work!
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
@gerhardb1227 Buenos días. Muy interesante su enfoque.
Entiendo que sólo hace una sustitución de variable.
Lo voy a intentar. Gracias.
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
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.
Hi, at the beginning of solution the obtained domain is incomplete. You should also solve 4-x>=0. So the domain is 1=
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
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.
THANKS PROFESOR !!!, VERY INTERESTING !!!!
You are welcome! Thank you very much!!
He sure took a loooong way to get there
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.
It's a very slippery slope to victory. if you take a wrong turn, you will find yourself in a dead end
Икс в квадрате минус один-это же (х-1)(х+1)😊
Precioso ejercicio.
Thank you very much 👍👍
by the way, perfect square is not the same as difference of squares
Много лишних шагов с преобразованиями
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.
The result is correct... x=1.25
Pl write what you are telling
Mi fa male la testa seguirti.😢😢😢😢
In your solution, there is an error in solving (x+1)(×-1)>=0. You forgot (x+1)=0 AND B>=0) OR (A
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.
Какой же университет закончил наш профессор, если на такое примитивное уравнение он потратил 25 (!) минут?
@@user-it6fh7hy6t Сам удивляюсь :)
Well, so did I, but these are screening problems. People who don't solve a whole bunch of these really fast do not qualify.
All of these are screening problems from timed tests. The real problems are multi-step projects involving informal proofs.
asnwer=3 isit
Профессор! Нормальный учитель распишет такое уравнение за 2 минуты,даже отвлекаясь на чай.
This is penise exercise. I don't understand
Much too complicated - square the original equation twice
It is more complicated like that and even the. you will still have some square roots left in the equation.
无聊
No need to write down that much