document.write( "Question 814491: There are numbers a, b, c, and d where the sum of any one of them and the product of the other three is equal to 2. If a>b, a>c, and a>d, find a. \n" ); document.write( "
Algebra.Com's Answer #490366 by Edwin McCravy(20055) $\"\"$ $\"About$
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document.write( "(1)   a + bcd = 2\r\n" );
document.write( "(2)   b + acd = 2\r\n" );
document.write( "(3)   c + abd = 2\r\n" );
document.write( "(4)   d + abc = 2\r\n" );
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document.write( "Subtract eq (2) from eq (1)\r\n" );
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document.write( "a - b + bcd - acd = 0\r\n" );
document.write( "a - b - acd + bcd = 0\r\n" );
document.write( "  (a-b) - cd(a-b) = 0\r\n" );
document.write( "      (a-b)(1-cd) = 0\r\n" );
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document.write( "Since a>b, a-b>0\r\n" );
document.write( "so         1-cd = 0         \r\n" );
document.write( "            -cd = -1\r\n" );
document.write( "             cd = 1\r\n" );
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document.write( "Subtract eq (3) from eq (1)\r\n" );
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document.write( "a - c + bcd - abd = 0\r\n" );
document.write( "a - c - abd + bcd = 0\r\n" );
document.write( "  (a-c) - bd(a-c) = 0\r\n" );
document.write( "      (a-c)(1-bd) = 0\r\n" );
document.write( "Since a>c, a-c>0\r\n" );
document.write( "so         1-bd = 0         \r\n" );
document.write( "            -bd = -1\r\n" );
document.write( "             bd = 1\r\n" );
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document.write( "Subtract eq (4) from eq (1)\r\n" );
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document.write( "a - d + bcd - abc = 0\r\n" );
document.write( "a - d - abc + bcd = 0\r\n" );
document.write( "  (a-d) - bc(a-d) = 0\r\n" );
document.write( "      (a-d)(1-bc) = 0\r\n" );
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document.write( "Since a>d, a-d>0\r\n" );
document.write( "so         1-bc = 0         \r\n" );
document.write( "            -bc = -1\r\n" );
document.write( "             bc = 1\r\n" );
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document.write( "So we have \r\n" );
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document.write( "cd = 1\r\n" );
document.write( "bd = 1\r\n" );
document.write( "bc = 1\r\n" );
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document.write( "cd = bd = 1, so c = b\r\n" );
document.write( "cd = bc = 1, so d = b\r\n" );
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document.write( "So b = c = d\r\n" );
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document.write( "a + bcd = 2\r\n" );
document.write( "a + bbb = 2\r\n" );
document.write( " a + b³ = 2\r\n" );
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document.write( "b + acd = 2\r\n" );
document.write( "b + abb = 2\r\n" );
document.write( "b + ab² = 2\r\n" );
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document.write( "So we have the system\r\n" );
document.write( " a + b³ = 2\r\n" );
document.write( "b + ab² = 2\r\n" );
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document.write( "Solve the 1st for a\r\n" );
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document.write( "a = 2 - b³\r\n" );
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document.write( "Substitute in the 2nd eq\r\n" );
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document.write( "     b + (2-b³)b² = 2\r\n" );
document.write( "     b + 2b² - b⁵ = 2\r\n" );
document.write( "-b⁵ + 2b² + b - 2 = 0\r\n" );
document.write( " b⁵ - 2b² - b + 2 = 0\r\n" );
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document.write( "Possible rational roots are ±1,±2\r\n" );
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document.write( "1|1 0 0 -2 -1  2\r\n" );
document.write( " |  1 1  1 -1 -2 \r\n" );
document.write( "  1 1 1 -1 -2  0\r\n" );
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document.write( "(b-1)(b⁴+b³+b²-b-2)=0\r\n" );
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document.write( "1|1 1 1 -1 -2\r\n" );
document.write( " |  1 2  3  2\r\n" );
document.write( "  1 2 3  2  0\r\n" );
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document.write( "(b-1)(b-1)(b³+2b²+3b+2)=0\r\n" );
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document.write( "-1|1  2  3  2\r\n" );
document.write( "  |  -1 -1 -2 \r\n" );
document.write( "   1  1  2  0\r\n" );
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document.write( "(b-1)(b-1)(b+1)(b²+b+2)=0\r\n" );
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document.write( "The last factor does not yield real\r\n" );
document.write( "roots.\r\n" );
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document.write( "So the real roots are 1 and -1\r\n" );
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document.write( "b can't be 1 for\r\n" );
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document.write( " a + b³ = 2\r\n" );
document.write( " a + 1³ = 2\r\n" );
document.write( "  a + 1 = 2\r\n" );
document.write( "      a = 1\r\n" );
document.write( "which contradicts a>b\r\n" );
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document.write( "Therefore b=c=d=-1\r\n" );
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document.write( "    a + b³ = 2\r\n" );
document.write( " a + (-1)³ = 2\r\n" );
document.write( "     a - 1 = 2\r\n" );
document.write( "         a = 3\r\n" );
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document.write( "Edwin
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