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Given Radical Real Product, Find the Square of Sum (Posted on 2023-05-25) Difficulty: 3 of 5
Each of x and y is a real number that satisfy this equation:

• {x+√(1+x2)}{y+√(1+y2)}= 1

Determine the value of (x+y)2

  Submitted by K Sengupta    
Rating: 5.0000 (1 votes)
Solution: (Hide)
Let u = x+√(1+x2)
Substituting this in the given equation, we obtain:
u(y+√(1+ y2)=1
Or, √(1+y2)= 1/u-y
Or, 1+y2= (1/u-y)2 = 1/u2 -2y/u+y2
Or, 2y/u = 1/u2 -1
Or, y = 1/2(1/u -u)

Now, u= x+ √(1+x2= (x-√(1+x2)/(x+√(1+x2)* (x-√(1+x2)= √(1+ x2)-x
So, 1/u = -x+ √(1+ x2 ), and:
u = x+√(1+ x2)

Hence, y = 1/2(1/u-u)= 1/2(-x + √(1+ x2-(x + √(1+ x2)
=(1/2)*(-2x) = -x
=> x+y=0 => (x+y)2 =0

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionSolutionBrian Smith2023-05-25 10:00:41
Some ThoughtsSpoiler?Steve Herman2023-05-25 09:30:22
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