If $(x + \sqrt{x^{2} + 1}) (y + \sqrt{y^{2} + 1}) = 1$ $(x+sqrt(x^2+1))(y+sqrt(y^2+1))=1$ what is $x + y$ $x+y$ ?

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If $(x + \sqrt{x^{2} + 1}) (y + \sqrt{y^{2} + 1}) = 1$ $(x+sqrt(x^2+1))(y+sqrt(y^2+1))=1$ what is $x + y$ $x+y$ ?

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Answer 1 · 2016-09-03T20:02:16

$x + y = 0$ $x+y = 0$

Explanation:

Let $y = - x$ $y = -x$

Then:

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

$(x + \sqrt{x^{2} + 1}) (y + \sqrt{y^{2} + 1}) = (x^{2} + 1) - x^{2}$ $color(white)((x+sqrt(x^2+1))(y+sqrt(y^2+1))) = (x^2+1)-x^2$

$(x + \sqrt{x^{2} + 1}) (y + \sqrt{y^{2} + 1}) = 1$ $color(white)((x+sqrt(x^2+1))(y+sqrt(y^2+1))) = 1$

So $x + y = 0$ $x+y = 0$ always results in $(x + \sqrt{x^{2} + 1}) (y + \sqrt{y^{2} + 1}) = 1$ $(x+sqrt(x^2+1))(y+sqrt(y^2+1)) = 1$

To see that this is the only possible value of $x + y$ $x+y$ , note that the function $f (x) = x + \sqrt{x^{2} + 1}$ $f(x) = x + sqrt(x^2+1)$ is strictly monotonic increasing.

So if $x_{1} + \sqrt{x_{1}^{2} + 1} = x_{2} + \sqrt{x_{2}^{2} + 1}$ $x_1+sqrt(x_1^2+1) = x_2+sqrt(x_2^2+1)$ then $x_{1} = x_{2}$ $x_1 = x_2$

Answer 2 · 2016-09-03T20:09:32

$x + y = 0$ $x+y=0$

Explanation:

If $x + \sqrt{x^{2} + 1} = δ$ $x+sqrt(x^2+1) = delta$ then $y + \sqrt{y^{2} + 1} = \frac{1}{δ}$ $y+sqrt(y^2+1) = 1/delta$

Solving

${(x - δ)}^{2} = x^{2} + 1$ $(x-delta)^2=x^2+1$ and

${(y - \frac{1}{δ})}^{2} = y^{2} + 1$ $(y-1/delta)^2=y^2+1$

we obtain

$x = \frac{δ^{2} - 1}{2 δ}$ $x = (delta^2-1)/(2delta)$ and
$y = - \frac{δ^{2} - 1}{2 δ}$ $y =- (delta^2-1)/(2delta)$

then

$x + y = 0$ $x+y=0$

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If $(x + \sqrt{x^{2} + 1}) (y + \sqrt{y^{2} + 1}) = 1$ $(x+sqrt(x^2+1))(y+sqrt(y^2+1))=1$ what is $x + y$ $x+y$ ?

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If $(x + \sqrt{x^{2} + 1}) (y + \sqrt{y^{2} + 1}) = 1$ $(x+sqrt(x^2+1))(y+sqrt(y^2+1))=1$ what is $x + y$ $x+y$ ?