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12 Oct 2020, 03:00
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Difficulty:
65%
(hard)
Question Stats:
61% (02:21) correct
39%
(02:44)
wrong
based on 225
sessions
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If \((x + \sqrt{1 + x^2})(y + \sqrt{1 + y^2}) = 1\), what is the value of \((x + y)^2\)?
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12 Oct 2020, 06:27
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\((x + √(1 + x^2))(y + √(1 + y^2))=1\)
put x=0 and y=0
\((x + √(1 + x^2))(y + √(1 + y^2)) = (0+1)*(0+1) =1\)
\((x+y)^2 = 0\)
put x=0 and y=0
\((x + √(1 + x^2))(y + √(1 + y^2)) = (0+1)*(0+1) =1\)
\((x+y)^2 = 0\)
Bunuel
General Discussion
12 Oct 2020, 12:19
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nick1816's solution is perfect, and is easily the fastest way to do the problem. You can also notice that if y = -x, the left side becomes (writing the sums in the opposite order to make the difference of squares pattern more obvious)
(√(1 + x^2) + x)(√(1 + x^2) - x) = 1 + x^2 - x^2 = 1
so any time y = -x, the equation will be true, and in that case (x + y)^2 = (x + (-x))^2 = 0.
(√(1 + x^2) + x)(√(1 + x^2) - x) = 1 + x^2 - x^2 = 1
so any time y = -x, the equation will be true, and in that case (x + y)^2 = (x + (-x))^2 = 0.
