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Algebra Level 2

Real numbers x x and y y satisfy x 1 1 x = ( x + y ) 2 \sqrt{x-1}-\sqrt{1-x}=(x+y)^2 . Then the value of x y x-y is

1 -1 2 2 3 3 1 1

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2 solutions

Sandeep Bhardwaj
Aug 22, 2014

As per given equation on L.H.S. :-

x 1 1 x \sqrt{x-1}-\sqrt{1-x}

for x 1 \sqrt{x-1} and 1 x \sqrt{1-x} to be defined , the only permissible value of x is 1.

i.e. DOMAIN of x 1 1 x \sqrt{x-1}-\sqrt{1-x} is {1}.

Putting x=1 in the given equation , we get y=-1

so x y = 1 ( 1 ) = 2 x-y=1-(-1)=\boxed{2}

. .
Feb 21, 2021

Only x = 1 x = 1 , and y = 1 y = -1 satisfies because 1 1 1 1 = ( 1 1 ) 2 0 0 = 0 2 0 0 = 0 \sqrt { 1 - 1 } - \sqrt { 1 - 1 } = ( 1 - 1 ) ^ { 2 } \Rightarrow \sqrt { 0 } - \sqrt { 0 } = 0 ^ { 2 } \Rightarrow 0 - 0 = 0 .

So the answer is 1 ( 1 ) = 2 1 - ( -1 ) = \boxed { \boxed { \boxed { \boxed { \boxed { 2 } } } } } .

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