document.write( "Question 1209723: Let p, q, r, and s be the roots of g(x) = 3x^4 - 8x^3 + 5x^2 + 2x - 17 - 2x^4 + 10x^3 + 11x^2 + 18x - 14.
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Algebra.Com's Answer #850025 by ikleyn(52781) $\"\"$ $\"About$

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\n" ); document.write( "Let p, q, r, and s be the roots of g(x) = 3x^4 - 8x^3 + 5x^2 + 2x - 17 - 2x^4 + 10x^3 + 11x^2 + 18x - 14.
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document.write( "Reduce the polynomial to the standard form, combining like terms\r\n" );
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document.write( "    3x^4 - 8x^3 + 5x^2 + 2x - 17 - 2x^4 + 10x^3 + 11x^2 + 18x - 14 = x^4 + 2x^3 + 16x^2 + 20x - 31.\r\n" );
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document.write( "Notice that \r\n" );
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document.write( "    p^2 + q^2 + r^2 + s^2 = (p + q + r + s)^2 - 2*(pq + pr +ps + qr + qs + rs)   (standard decomposition for the square of a sum)\r\n" );
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document.write( "Also notice that the leading coefficient of the polynomial standard form at x^4 is 1.\r\n" );
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document.write( "Due to Vieta's theorem\r\n" );
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document.write( "    pq + pr + ps + qr + qs + rs = 16  (the sum of in-pairs products of the roots is the coefficient at x^2)\r\n" );
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document.write( "    p + q + r + s =  -2               (the sum of the roots is the coefficient at x^3 with the opposite sign).\r\n" );
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document.write( "Therefore,\r\n" );
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document.write( "    p^2 + q^2 + r^2 + s^2 = (-2)^2 - 2*16 = 4 - 32 = -28.\r\n" );
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document.write( "ANSWER.  p^2 + q^2 + r^2 + s^2 = -28.   \r\n" );
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document.write( "         (by the way, since the sum of squares is negative, it means that some roots are complex numbers).\r\n" );
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