document.write( "Question 408299: Given one roots of x^2+px+8=0 is two times of the other. Find the roots and p. \n" ); document.write( "
Algebra.Com's Answer #287703 by richard1234(7193)\"\" \"About 
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Interesting thing about this problem is that the roots and the value of p are not uniquely determined. You'll see why:\r
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\n" ); document.write( "\n" ); document.write( "Suppose the two roots are \"r\" and \"2r\". The product of the roots is 8 by Vieta's formulas, so \"2r%5E2+=+8\" --> \"r%5E2+=+4\" --> r = +/- 2. Hence, the roots are {2, 4} or {-2, -4}. Also, the sum of the roots is -p, so p = -6 or 6, respectively.\r
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\n" ); document.write( "\n" ); document.write( "Hence, the quadratic \"x%5E2+-+6x+%2B+8\" has roots {2, 4} and \"x%5E2+%2B+6x+%2B+8\" has roots {-2, -4}.
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