Algebra.Com's Answer #684042 by ikleyn(52781)![\"\"](\"/images/stars/stars-13.gif\")  
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document.write( "Without solving the given equation, find an equation whose roots are the squares of the roots of x^2 + 4x + 2 = 0.
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document.write( "The trick and the focus is to find the second equation without solving the first.\r
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document.write( "In this sense the solution by \"josgarithmetic\" is \"out of the target\".\r
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document.write( "I will show you how to strike EXACTLY to the target.\r
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document.write( "It is about the Vieta's theorem, whether it is included or not included to the school math curriculum. \r\n" );
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document.write( "Let \"s\" and \"t\" be the roots of the quadratic polynomial\r\n" );
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document.write( "p(x) =  \r\n" );
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document.write( "with the leading coefficient 1 at  . Then the Vieta's theorem says:\r\n" );
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document.write( "s + t = -p and s*t = q.\r\n" );
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document.write( "For the given equation \r\n" );
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document.write( " =  \r\n" );
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document.write( "it means that if \"s\" and \"t\" are its roots, then\r\n" );
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document.write( "s + t = -4, (1) and\r\n" );
document.write( "s*t = 2. (2)\r\n" );
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document.write( "Next, if (1) and (2) are held, then\r\n" );
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document.write( " =  =  = 16; hence,  = 16 - 2s*t = 16 - 2*2 = 12, and\r\n" );
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document.write( " =  =  = 4.\r\n" );
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document.write( "Hence, by applying the Vieta;s theorem once again, you see that  and  are the roots of the polynomial\r\n" );
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document.write( "g(x) =  .\r\n" );
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document.write( "It is the answer to the problem's question. The problem is solved !!\r\n" );
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document.write( "Notice, we get the answer without solving the original equation.\r\n" );
document.write( "Exactly as assigned by the condition.\r\n" );
document.write( "All we did we manipulated with coefficients and used the Vieta's theorem.\r\n" );
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document.write( "On the way you learned about Vieta's theorem for quadratic equations and polynomials.\r\n" );
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document.write( "Happy learning !!\r
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