Prove $x=-y$: A Math Challenge

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SUMMARY

The mathematical challenge presented involves proving that if \(x\) and \(y\) are real numbers satisfying the equation \((\sqrt{y^{2} - x} - x)(\sqrt{x^{2} + y} - y) = y\), then \(x\) must equal \(-y\). Substituting \(-y\) for \(x\) in the left-hand side simplifies the expression to equal \(y\), confirming that \(x = -y\) is indeed a solution. The identity \((a+b)(a-b) = (a^2 - b^2)\) is crucial for this simplification.

PREREQUISITES
  • Understanding of real numbers and their properties
  • Familiarity with square roots and algebraic manipulation
  • Knowledge of mathematical identities, specifically \((a+b)(a-b) = a^2 - b^2\)
  • Basic skills in solving equations
NEXT STEPS
  • Study algebraic identities and their applications in proofs
  • Learn about the properties of square roots in real number equations
  • Explore methods for proving mathematical statements and identities
  • Practice solving similar equations involving real numbers and square roots
USEFUL FOR

Mathematics students, educators, and anyone interested in algebraic proofs and the properties of real numbers.

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Let $x, y$ be real numbers such that
$$(\sqrt{y^{2} - x\,\,}\, - x)(\sqrt{x^{2} + y\,\,}\, - y)=y.$$
Prove $x=-y$.

Any suggestion would be appreciated.
 
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Another way of stating the question is:

Show that x=-y is a solution to:

\left[\sqrt{y^2-x}-x \right]\left[\sqrt{x^2+y}-y\right]=y

Writing it is this way it is more obvious that all you have to do is substitute (-y) for each of the x's in the left hand side and then simplify to show that the left hand side is equal to y.

To do this it will be helpful to remember the identity:
(a+b)(a-b)=(a^2-b^2)
 
Kiwi said:
Another way of stating the question is:

Show that x=-y is a solution to:

\left[\sqrt{y^2-x}-x \right]\left[\sqrt{x^2+y}-y\right]=y
No, this is the converse of what the original question is asking.
 

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