document.write( "Question 1073256: If p and q are the roots of the equation
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Algebra.Com's Answer #688146 by ikleyn(52782) $\"\"$ $\"About$

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document.write( "This assignment is for those who firmly knows the Vieta's theorem and Vieta's formulas.\r\n" );
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document.write( "    For the second degree polynomial (quadratic)  P(x)=   roots  and   of the equation  P(x)=0 satisfy\r\n" );
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document.write( "     +  =    and    = .     (1)     (see Wikipedia, this article).  \r\n" );
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document.write( "Concretely, for the given equation   = 0  its roots p and q satisfy \r\n" );
document.write( "     p + q =  and p*q = -2.     (2)\r\n" );
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document.write( "We can state it based in Vieta's theorem even without making explicit calculations of the roots.\r\n" );
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document.write( "Now, they want we construct the quadratic equation/polynomial with the roots  and .\r\n" );
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document.write( "1.  Then the coefficient at x of this polynomial, based on the Vieta's theorem,  must be the opposite number to\r\n" );
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document.write( "         +  =  -  =  - .       (3)\r\n" );
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document.write( "        From (2), you have  =  =  =  = .\r\n" );
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document.write( "        Therefore, we can continue and complete (3) in this way\r\n" );
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document.write( "         +  =  -  =  =  =  = .\r\n" );
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document.write( "        So, the opposite number to it,  is the coefficient at x in the polynomial under the question.\r\n" );
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document.write( "2.  Next step we should calculate the product . to determine the constant term of our polynomial.  The product is \r\n" );
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document.write( "        . =  -  -  +  =  - .\r\n" );
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document.write( "        Now, replace pq by -2 and replace   by   =  =  =  =  = .\r\n" );
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document.write( "        You will get  . = ((-2) + 1) -  =  = .\r\n" );
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document.write( "Thus your equation under the question is   = 0.\r\n" );
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document.write( "Or, if you want to have it with integer coefficients, multiply everything by 16, and you will get\r\n" );
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document.write( "         = 0.\r\n" );
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\n" ); document.write( "\n" ); document.write( "Using this approach, you have a priviledge of making calculations with rational numbers.\r
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\n" ); document.write( "\n" ); document.write( "By doing it using another approach with explicit irrationalities, you will be forced to work with radicals.\r
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\n" ); document.write( "\n" ); document.write( "It is the advantage of using Vieta's theorem.\r
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\n" ); document.write( "To extend your horizon:\r
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document.write( "     The way I showed here, is THE ONLY WAY to solve the problems like this one.\r\n" );
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document.write( "     The problem is especially designed and intended to teach you this method.\r\n" );
document.write( "     It is not my fantasy. It is the way you should know and the approach you must follow when solving such problems.\r\n" );
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document.write( "     It may happen you will have a desire to find the roots of the original equations and then manipulate with them.\r\n" );
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document.write( "     It is NOT THE WAY they want you follow in your solution.\r\n" );
document.write( "     Saying \"they\", I mean those who assigned you this problem.\r\n" );
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document.write( "     They DEFINITELY want to know, whether you know the method I showed you, and\r\n" );
document.write( "                                   whether you are able to use it.\r\n" );
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\n" ); document.write( "\n" ); document.write( "To see similar solved problems, look into the lessons\r
\n" ); document.write( "\n" ); document.write( "    - HOW TO evaluate expressions involving $\"%28x+%2B+1%2Fx%29\"$ , $\"%28x%5E2%2B1%2Fx%5E2%29\"$ , $\"%28x%5E3+%2B+1%2Fx%5E3%29\"$ and $\"%28x%5E5%2B1%2Fx%5E5%29\"$ \r
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\n" ); document.write( "\n" ); document.write( "Also, you have this free of charge online textbook in ALGEBRA-I in this site\r
\n" ); document.write( "\n" ); document.write( "    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.\r
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\n" ); document.write( "\n" ); document.write( "The referred lessons are the part of this online textbook under the topic \"Evaluation, substitution\".\r
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\n" ); document.write( "\n" ); document.write( "        Good luck and     H A P P Y     L E A R N I N G ! !\r
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