Japanese | Can you solve this ? | A Nice Math Olympiad Algebra Problem
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Published 2023-12-29
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All Comments (21)
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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
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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.
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Nice problem. Good solution
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There is a mistake a² >= 0, then a >= 0 or a<= 0 . With b² >= 2 then b>= sqrt(2) or b<= sqrt(2). But a=sqrt(x-1) and b=sqrt(x+1), with x>= 1, then a >= 0 and b >= sqrt(2). The solution is OK, but the reasoning is incorrect
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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
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The presenter makes the incorrect statement that "a squared - b squared is in the form of a perfect square" more than once.
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Lovely work!
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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
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He sure took a loooong way to get there
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Hi, at the beginning of solution the obtained domain is incomplete. You should also solve 4-x>=0. So the domain is 1=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.THANKS PROFESOR !!!, VERY INTERESTING !!!!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.It's a very slippery slope to victory. if you take a wrong turn, you will find yourself in a dead endPrecioso ejercicio.by the way, perfect square is not the same as difference of squaresИкс в квадрате минус один-это же (х-1)(х+1)😊The result is correct... x=1.25Condition 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.
