document.write( "Question 1163497: Question: Equation: x^3 + ax^2 +bx +a = 0 |a,b are real
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Algebra.Com's Answer #787607 by ikleyn(52794) $\"\"$ $\"About$

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document.write( "Since the coefficients of the polynomial are real numbers and one root is (2+i),  the other root is complex conjugate (2-i).\r\n" );
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document.write( "Let the third root is x  (it is clear that the third root is a real number).\r\n" );
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document.write( "Now use the Vieta's theorem.\r\n" );
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document.write( "It says that the product of the roots is the constant term with the opposite sign:\r\n" );
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document.write( "    (2+i)*(2-i)*x = -a,   or\r\n" );
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document.write( "    5x = -a.             (1)\r\n" );
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document.write( "The Vieta;s theorem also says that the sum of the three roots is equal to a coefficient at x^2 with the opposite sign:\r\n" );
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document.write( "    2+i + 2-i + x = -a,  or\r\n" );
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document.write( "    4 + x = -a.          (2)\r\n" );
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document.write( "So, you have this system of two equations (1) and (2).\r\n" );
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document.write( "Based on (1), replace \"-a\" in (2) by 5x.  You will get then\r\n" );
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document.write( "    4 + x = 5x\r\n" );
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document.write( "Hence, x= 1  and  a= -5x = -5.\r\n" );
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document.write( "To find the coefficient \"b\", apply the Vieta's theorem again.\r\n" );
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document.write( "It says that the coefficient b is equal to the sum of the three in-pairs products of the roots\r\n" );
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document.write( "    b = (2+i)*(2-i) + 1*(2+i) + 1*(2-i) = 5 + 2+i + 2-i = 5 + 4 = 9.\r\n" );
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document.write( "ANSWER.  a= -5,  b= 9.\r\n" );
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