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Posted (edited)

Looks to me more like a 4th order polynomial.

But only one of the four solutions to that 4th order polynomial works in the sense that the square root symbol means the principal value.

 

Hint: This one sensible solution has integer values of x and y.

Edited by D H
Posted

Here is my answer to the question:

 

 

A quick inspection of the two equations shows that (9,4) is a solution to that equation.

 

[math]Let: x=9, y=4[/math]

[math]\sqrt{9}+4=3+4=7[/math]

[math]9+\sqrt{4}=9+2=11[/math]

 

To be honest I didn't really do any algebraic work. I knew that x and y must be positive integer, and more than likely perfect squares. This means y must be less than 7. Since you only have to check the perfect squares y is either 4 or 1. Likewise, x must be less than 11, and since I am assuming x and y are perfect squares you only have to check 9, 4, and 1. Quickly looking at these possible solution sets gives one the answer that x=9, y=4.

 

 

Posted

Here is my answer to the question:

 

Indeed. The flaw in the problem is that the answer is too obvious before you formally solve it. Narrowing down to sensible guesses and using trial and error is a shortcut I've often used, but here there is exactly one sensible guess.

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