document.write( "Question 696524: in a quadratic equation x^2 + px + q =0 if sum of its roots are equal to thrice of their difference then proove that
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Algebra.Com's Answer #429077 by Edwin McCravy(20055)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "In this approach we don't actually find the roots as the other\r\n" );
document.write( "tutor did, although that method is quite correct.\r\n" );
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document.write( "x² + px + q = 0\r\n" );
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document.write( "Let the roots be r1 and r2.  \r\n" );
document.write( "Then the above factors as\r\n" );
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document.write( "(x - r1)(x - r2) = 0\r\n" );
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document.write( "or, upon \"FOIL\"ing that out:\r\n" );
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document.write( "x² - r2x - r1x + r1r2 = 0\r\n" );
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document.write( "x² - (r2+r1)x + r1r2 = 0\r\n" );
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document.write( "x² - (r1+r2)x + r1r2 = 0\r\n" );
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document.write( "comparing this to\r\n" );
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document.write( "x² + px + q = 0\r\n" );
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document.write( "p = -(r1+r2)\r\n" );
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document.write( "q = r1r2\r\n" );
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document.write( "Now we use the fact that the sum of its roots \r\n" );
document.write( "are equal to thrice their difference, which\r\n" );
document.write( "says:\r\n" );
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document.write( "r1+r2 = 3(r1-r2)  or r1+r2 = 3(r2-r1)\r\n" );
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document.write( "Both equations are technically necessary because we are not told \r\n" );
document.write( "which order we are to subtract the roots to find the difference.\r\n" );
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document.write( "r1+r2 = 3r1-3r2  or r1+r2 = 3r2-3r1\r\n" );
document.write( "  4r2 = 2r1      or   4r1 = 2r2\r\n" );
document.write( "  2r2 = r1       or   2r1 = r2\r\n" );
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document.write( "But since they are the same except for which root we call\r\n" );
document.write( "r1 and which we call r2, we can just take the first case.\r\n" );
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document.write( "Now we have these 3 equations:\r\n" );
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document.write( "p = -(r1+r2)\r\n" );
document.write( "q = r1r2\r\n" );
document.write( "2r2 = r1\r\n" );
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document.write( "Using the third equation, we substitute 2r2 for r1 in the\r\n" );
document.write( "first two equations:\r\n" );
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document.write( "p = -(2r2+r2)\r\n" );
document.write( "q = 2r2r2\r\n" );
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document.write( "Simplifying,\r\n" );
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document.write( "p = -3r2\r\n" );
document.write( "q = 2r2²\r\n" );
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document.write( "Solve the first one for r2:\r\n" );
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document.write( "r2 = \"p%2F%28-3%29\"\r\n" );
document.write( "r2 = \"-p%2F3\"\r\n" );
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document.write( "Substitute \"-p%2F3\" for r2 in q = 2r2²:\r\n" );
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document.write( "q = 2\"%28-p%2F3%29%5E2\"\r\n" );
document.write( "q = 2\"%28p%5E2%2F3%5E2%29\"\r\n" );
document.write( "q = 2\"%28p%5E2%2F9%29\"\r\n" );
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document.write( "Multiply both sides by 9\r\n" );
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document.write( "9q = 2p² \r\n" );
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document.write( "That's the same as\r\n" );
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document.write( "2p² = 9q\r\n" );
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document.write( "Edwin
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