Solving for $2012x^2-2010y^2+2011x-2011y-2011$ with $x,y\in R$

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In summary, the given equation is a quadratic equation with two variables and coefficients of 2012, -2010, 2011, and -2011. It can have an infinite number of solutions as long as x and y are real numbers and can be solved using various methods such as factoring, completing the square, or using the quadratic formula. The equation does not have imaginary solutions since it only contains real coefficients and variables. To better understand the equation, it can be graphed using a graphing calculator or plotting points on a coordinate plane to show its parabolic shape and intersections with the x and y axes.
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Albert1
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$x,y\in R$, satisfying $(x-\sqrt{x^2-2011})(y-\sqrt{y^2-2011})=2011$
find $2012x^2-2010y^2+2011x-2011y-2011=?$
 
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Albert said:
$x,y\in R$, satisfying $(x-\sqrt{x^2-2011})(y-\sqrt{y^2-2011})=2011$
find $2012x^2-2010y^2+2011x-2011y-2011=?$

My solution:
Note that we could rewrite the given equality as

$(x-\sqrt{x^2-2011})(y-\sqrt{y^2-2011})=2011$

$x-\sqrt{x^2-2011}=\dfrac{2011}{y-\sqrt{y^2-2011}}$

$x-\sqrt{x^2-2011}=\dfrac{2011(y+\sqrt{y^2-2011})}{(y-\sqrt{y^2-2011})(y+\sqrt{y^2-2011})}=y+\sqrt{y^2-2011}$

Note that for $x\ge \sqrt{2011}$, the function on the right is an increasing function and its starting point is at $(\sqrt{2011},\, \sqrt{2011})$ while the function on the left is a decreasing function with its starting point $(\sqrt{2011},\, \sqrt{2011})$, (whereas for the region where $x\le \sqrt{2011}$, the function on the right is an decreasing function and its starting point is at $(-\sqrt{2011},\, -\sqrt{2011})$ while the function on the left is an increasing function with its starting point $(-\sqrt{2011},\, -\sqrt{2011})$) therefore we can conclude the only solution the system has is when $x^2=2011=y^2$, or $x=y=\pm\sqrt{2011}$

Therefore,

$\begin{align*}2012x^2-2010y^2+2011x-2011y-2011&=(2012-2010)(2011)-2011\\&=2011(2-1)\\&=2011\end{align*}$
 

FAQ: Solving for $2012x^2-2010y^2+2011x-2011y-2011$ with $x,y\in R$

1. What is the equation $2012x^2-2010y^2+2011x-2011y-2011$?

The equation is a quadratic equation with two variables, $x$ and $y$, and coefficients of 2012, -2010, 2011, and -2011.

2. What are the possible solutions for the equation?

The equation can have an infinite number of solutions, as long as $x$ and $y$ are real numbers.

3. How do I solve for the values of $x$ and $y$ in the equation?

To solve for the values of $x$ and $y$, you can use various methods such as factoring, completing the square, or using the quadratic formula.

4. Can the equation have imaginary solutions?

No, since the equation only contains real coefficients and variables, the solutions must also be real numbers.

5. How can I graph the equation to better understand it?

To graph the equation, you can use a graphing calculator or plot points on a coordinate plane. The graph will show the parabolic shape of the equation and the points where it intersects the x and y axes.

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