document.write( "Question 772758: If one of the roots of the quadratic equation $\"x%5E2-%282k-5%29x%2Bk%5E2-5k%2B56%2F9=0\"$ is twice the other,show that one of the roots is $\"%282k-5%29%2F3\"$ .Hence find the values of the constant k.\r
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Algebra.Com's Answer #471100 by Edwin McCravy(20055) $\"\"$ $\"About$

You can put this solution on YOUR website!

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document.write( "x² - (2k-5)x + (k²-5k+56/9) = 0\r\n" );
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document.write( "Use these facts of quadratics with leading coefficient 1:\r\n" );
document.write( "For the quadratic equation x²+Ax+B=0\r\n" );
document.write( "The sum of the roots  is -A\r\n" );
document.write( "The product of the roots is B\r\n" );
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document.write( "Let r = the root that is NOT twice the other\r\n" );
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document.write( "Then the root which IS twice the other is 2r\r\n" );
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document.write( "Sum of roots = r+2r = 3r\r\n" );
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document.write( "Therefore \r\n" );
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document.write( "         3r = 2k-5\r\n" );
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document.write( "and       r = \r\n" );
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document.write( "That was the first thing you were to show.\r\n" );
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document.write( "Product of roots = (r)(2r) = 2r²\r\n" );
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document.write( "Therefore \r\n" );
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document.write( "        2r² = k²-5k+\r\n" );
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document.write( "So we have the system of equations:\r\n" );
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document.write( "1st:     3r = 2k-5\r\n" );
document.write( "2nd:    2r² = k²-5k+\r\n" );
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document.write( "Clear the 2nd equation of fractions:\r\n" );
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document.write( "       18r² = 9k²-45k+56\r\n" );
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document.write( "Square both sides of the 1st\r\n" );
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document.write( "        9r² = 4k²-20k+25\r\n" );
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document.write( "Multiply through by 2\r\n" );
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document.write( "       18r² = 8k²-40k+50\r\n" );
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document.write( "Set the expressions for 18x² equal:\r\n" );
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document.write( " 9k²-45k+56 = 8k²-40k+50\r\n" );
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document.write( "k² - 5k + 6 = 0\r\n" );
document.write( " (k-3)(k-2) = 0\r\n" );
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document.write( "k-3=0;  k-2=0\r\n" );
document.write( "  k=3;    k=2\r\n" );
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document.write( "So those are the 2 values of k, which is what was asked for.\r\n" );
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document.write( "We weren't asked for the roots, but we could find the roots of the given\r\n" );
document.write( " equation very easily:\r\n" );
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document.write( "for k = 3, substitute in\r\n" );
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document.write( "3r = 2k-5\r\n" );
document.write( "3r = 2(3)-5\r\n" );
document.write( "3r = 6-5\r\n" );
document.write( "3r = 1\r\n" );
document.write( " r = \r\n" );
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document.write( "So for k = 3, one root is x =  and the other is x =  \r\n" );
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document.write( "for k = 2, substitute in\r\n" );
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document.write( "3r = 2k-5\r\n" );
document.write( "3r = 2(2)-5\r\n" );
document.write( "3r = 4-5\r\n" );
document.write( "3r = -1\r\n" );
document.write( " r = \r\n" );
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document.write( "So for k = 2, one root is x =  and the other is x = \r\n" );
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document.write( "Edwin

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