Question

If the real number x , y satisfy the condition

\big( x+\sqrt{1+x^2}\big) \big( y+\sqrt{1+y^2} \big) =1 ,

verify the value of the expression (x+y)^2 is equal to 0.

Collected in the board: Algebraic proof questions

Steven Zheng Steven Zheng posted 1 month ago


Answer

Let

n = x+\sqrt{1+x^2}
(1)

The condition expression is transformed to ,

n(y+\sqrt{1+y^2})=1


y+\sqrt{1+y^2} = \dfrac{1}{n}


\sqrt{1+y^2} = \dfrac{1}{n}-y


Square the both sides of the equation


\Big( \sqrt{1+y^2}\Big) ^2 = \Big( \dfrac{1}{n}-y\Big) ^2


1+y^2 = \dfrac{1}{n^2}- \dfrac{2y}{n}+y^2
(2)

\dfrac{1}{n^2}- \dfrac{2y}{n}=1


-2y = n-\dfrac{1}{n}


y = \dfrac{1}{2}(\dfrac{1}{n}-n)
(3)

Replace n with (1)

\dfrac{1}{n}-n

=\dfrac{1}{ x+\sqrt{1+x^2}}-( x+\sqrt{1+x^2})

=\dfrac{ x-\sqrt{1+x^2}}{( x+\sqrt{1+x^2})( x-\sqrt{1+x^2})} -( x+\sqrt{1+x^2})

=\dfrac{ x-\sqrt{1+x^2}}{-1} -( x+\sqrt{1+x^2})

=-2x


\therefore y = -x


(x+y)^2=0

Steven Zheng Steven Zheng posted 1 month ago

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