document.write( "Question 1111112: If (x+ysqrt3)^2 = 48+ 24sqrt3, find the values of x and y, given that they are rational. \n" ); document.write( "

Algebra.Com's Answer #726120 by ikleyn(52793) $\"\"$ $\"About$

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document.write( "x^2 + 3y^2 = 48\r\n" );
document.write( "2xy        = 24    ====>  xy =  = 12.\r\n" );
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document.write( "x =    x^2 = \r\n" );
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document.write( " + +  = 48\r\n" );
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document.write( "144 + 3y^4 = 48y^2\r\n" );
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document.write( "3y^4 - 48y^2 + 144 = 0\r\n" );
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document.write( "y^4 - 16y^2 + 48 = 0\r\n" );
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document.write( " =  = .\r\n" );
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document.write( "1)  y^2 =  = 12  ====>  y is irrational  ===>  the solution does not work;\r\n" );
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document.write( "2)  y^2 =  = 4  ====>  y = +/-2  ====>  x =  = +/- 6.\r\n" );
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document.write( "Answer.  Two solutions are  (x,y) = (6,2);  (x,y) = (-6,-2).\r\n" );
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