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Mistake in the beginning in the Domain discussion and lack of elegance and compactness in the solving. Much sharper as follow : It is clear that x>=1 spans the Domain of {E} : x+✔️(x-1}+ ✔️(x+1}+ ✔️{(x-1)(x+1)}=4. Define a= ✔️(x-1} and b= ✔️(x+1} . Then aa+bb=2x and bb-aa=2=(b-a)(b+a) rechape {E} in a and b variables : (a+b)^2+2(a+b)=8. Which factorises by completing the square in : (a+b+1)^2-3^2=0. Giving only a+b=2. Thus b-a=1. Which linear system solves in : a=1/2 and b=3/2. Giving one valid final solution : x=5/4. QED
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
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
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
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.
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
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.
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.
He took a long route. If he looked at the LHS of the original equation, he could've studied the middle 2 terms. The product is the 4th term. If he squared the sum of the 2 terms, he'll note that the result is twice the sum of the 1st and 4th terms. Thus, he can denote the sum as S, and have: S²/2 + S = 4 S²/2 + S - 4 = 0 S = (-1 ± √(1+8))/1 = -1 ± 3 = 2, -4 Thus, we have: √(x-1) + √(x+1) = 2 √(x+1) = 2 - √(x-1) x+1 = 4 - 4*√(x-1) + x-1 √(x-1) = 1/2 x-1 = 1/4 x = 5/4 There's no reason to add 1 and subtract 1 to generate 2 other equations.
