document.write( "Question 1089561: If x^2 + y^2 =2 and xy = 1, find x^2-y^2 \n" ); document.write( "

Algebra.Com's Answer #703898 by ikleyn(52864) $\"\"$ $\"About$

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\n" ); document.write( "If x^2 + y^2 =2 and xy = 1, find x^2-y^2
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document.write( "If  \r\n" );
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document.write( "x^2 + y^2 =2      (1)    and \r\n" );
document.write( "xy = 1,           (2)\r\n" );
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document.write( "then  x = ,   =   and, after substituting it into (1), you get this equation for the single unknown y\r\n" );
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document.write( " +  = 2,    which is the same as\r\n" );
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document.write( " - 2 +  = 0,   or\r\n" );
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document.write( " = 0,   which implies\r\n" );
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document.write( "y = 1/y   and then   y^2 = 1;  finally,   y = 1   OR  y= -1.\r\n" );
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document.write( "If y = 1, then, obviously x = +/-1;   If y = -1, then, again, x = +/-1.\r\n" );
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document.write( "Taking into account that xy = 1, you can conclude that the solutions to (1),(2) are these two pairs (two points):  (1,1)   and   (-1,-1).\r\n" );
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document.write( "And you can easily check that in both cases   -  = 0.\r\n" );
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document.write( "Answer.  If x^2 + y^2 =2 and xy = 1,  then  x^2-y^2 = 0.\r\n" );
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