document.write( "Question 1075157: Please help me solve this problem :
\n" ); document.write( "Let M and N be the root of an equation x^2+ax+b=0 and
\n" ); document.write( "let O and P be the roots of x^2+cx+d=0. Express
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Algebra.Com's Answer #689864 by ikleyn(52794)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "1.  (M-O)*(N-O) = \"MN+-+ON+-OM+%2BO%5E2\" \r\n" );
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document.write( "                = \"b+-+O%2A%28M%2BN%29+%2B+O%5E2\"          (I replaced MN by b based on Vieta's formulas)\r\n" );
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document.write( "                = \"b+%2B+cO+%2BO%5E2\"     (1)      (I replaced M+N by -c  based on Vieta's formulas)\r\n" );
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document.write( "2.  (M-P)*(N-P) = \"MN+-+PN+-PM+%2BP%5E2\" \r\n" );
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document.write( "                = \"b+-+P%2A%28M%2BN%29+%2B+P%5E2\"          (I replaced MN by b based on Vieta's formulas)\r\n" );
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document.write( "                = \"b+%2B+cP+%2B+P%5E2\"     (2)      (I replaced M+N by -c  based on Vieta's formulas)\r\n" );
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document.write( "3.  (M-O)*(N-O)*(M-P)*(N-P) = \"%28b+%2B+cO+%2BO%5E2%29%2A%28b+%2B+cP+%2BP%5E2%29\" =  (I used here (1) and (2) )\r\n" );
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document.write( "    =  = \r\n" );
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document.write( "    =  = \r\n" );
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document.write( "    =  = \r\n" );
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document.write( "    = \"b%5E2+-+bc%5E2+%2B+b%2A%28%28-c%29%5E2-2d%29+%2B+c%5E2d+-+c%5E2d+%2B+d%5E2\"\r\n" );
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document.write( "         ( I replaced here P+O by -c, \"P%5E2+%2B+O%5E2\" by \"%28P%2BO%29%5E2-2PO%29\", and PO by d. )\r\n" );
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document.write( "    = \"b%5E2+-+bc%5E2+%2B+b%2A%28c%5E2-2d%29+%2B+d%5E2\" = \"b%5E2+-+2bd+%2B+d%5E2\" = \"%28b-d%29%5E2\".     ( !!! )\r\n" );
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\n" ); document.write( "\n" ); document.write( "Check my Math. I could make a mistake.\r
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\n" ); document.write( "\n" ); document.write( "But the major idea is VERY CLEAR:\r
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document.write( "     make multiplications and use everywhere the Vieta's formulas for quadratic polynomials: \r\n" );
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document.write( "         the sum of the roots is the coef. at x with the opposite sign; the product of the roots is the constant term.\r\n" );
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