document.write( "Question 1138927: The roots of x²+kx+6=0 are each 5 less than the roots of x²−kx+6=0. Find k.\r
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Algebra.Com's Answer #756707 by Edwin McCravy(20056) $\"\"$ $\"About$

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The roots of x²+kx+6 = 0 are each 5 less than the roots of x²−kx+6 = 0. Find k.
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document.write( "Let the roots of the second equation x²-kx+6 = 0 be r and s. Then since\r\n" );
document.write( "the leading coefficient is 1, the sum of the roots equals the coefficient\r\n" );
document.write( "of x with the sign changed, so,\r\n" );
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document.write( "r+s = k\r\n" );
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document.write( "Then the roots of the first equation x²+kx+6 = 0 are r-5 and s-5. Then since\r\n" );
document.write( "the leading coefficient is 1, the sum of the roots equals the coefficient\r\n" );
document.write( "of x with the sign changed, so,\r\n" );
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document.write( "r-5 + s-5 = -k, which simplifies to\r\n" );
document.write( "r+s-10 = -k\r\n" );
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document.write( "So we have the system:\r\n" );
document.write( "r+s = k\r\n" );
document.write( "r+s-10 = -k\r\n" );
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document.write( "Substitute k for r+s in the second:\r\n" );
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document.write( "k-10 = -k\r\n" );
document.write( "  2k = 10\r\n" );
document.write( "   k = 5\r\n" );
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document.write( "Checking:\r\n" );
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document.write( "Solve:\r\n" );
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document.write( "   x²+5x+6 = 0           x²-5x+6 = 0\r\n" );
document.write( "(x+3)(x+2) = 0        (x-3)(x-2) = 0   \r\n" );
document.write( " x+3=0; x+2=0          x-3=0; x-2=0 \r\n" );
document.write( "  x=-3;   x=-2           x=3;   x=2\r\n" );
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document.write( "the root -3 of the first equation is 5 less than the root 2 of the\r\n" );
document.write( "second equation, since -3 = 2-5.  Also:\r\n" );
document.write( "the root -2 of the first equation is 5 less than the root 3 of the\r\n" );
document.write( "second equation, since -2 = 3-5. \r\n" );
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document.write( "Edwin

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