document.write( "Question 1152368: √x+√(1-x)+√(x(1-x))=1 \n" ); document.write( "

Algebra.Com's Answer #774575 by ikleyn(52781) $\"\"$ $\"About$

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document.write( " +  +  = 1.      (1)\r\n" );
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document.write( "The domain, where all included functions are defined, is the segment  [0,1].\r\n" );
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document.write( "Two obvious solutions to the given equation in this domain are x= 0  and  x= 1.\r\n" );
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document.write( "Below I will show that the given equation HAS NO other solutions.\r\n" );
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document.write( "Indeed, let  0 < x < 1.\r\n" );
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document.write( "Then         is defined and is positive number   > 0.\r\n" );
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document.write( "Similarly,   is defined and is positive number   > 0.\r\n" );
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document.write( "    For any two real positive numbers \"a\" and \"b\" the following inequality is valid\r\n" );
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document.write( "        a + b > .    (2)\r\n" );
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document.write( "    To prove it, square both sides. You will get\r\n" );
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document.write( "        a^2 + 2ab + b^2 > a^2 + b^2,\r\n" );
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document.write( "    which is valid for all positive \"a\" and \"b\".\r\n" );
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document.write( "Now apply the inequality (2) for  a=   and  b= .  You will get\r\n" );
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document.write( "     +  >  =  =  = 1.\r\n" );
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document.write( "Thus,  the sum   +   at  0 < x < 1  is just greater than 1.\r\n" );
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document.write( "With the added positive addend  ,  the sum   +  +   is just even more than 1.  \r\n" );
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document.write( "Therefore,  the sum   +  +   can not be equal to 1  at  0 < x < 1.\r\n" );
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document.write( "Thus, it is  PROVED that the given equation has no solutions inside the segment [0,1].  \r\n" );
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document.write( "So, the endpoints  x= 0  and  x= 1 are the only solutions.\r\n" );
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\n" ); document.write( "\n" ); document.write( "Solved.\r
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