document.write( "Question 1136685: : If one root of the equation x^2-px+20=0 is four, while the equation x^2-qx+p=0 has equal roots then, then a possible value of q is
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Algebra.Com's Answer #754486 by ikleyn(52781) $\"\"$ $\"About$

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document.write( "The condition says that  x^2 - px + 20 = 0  has the root 4.\r\n" );
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document.write( "Then the other root is   = 5,  according to Vieta's theorem.\r\n" );
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document.write( "It implies (via Vieta's theorem, again) that  p  is equal to the sum of the roots  5+4 = 9:  p = 9.\r\n" );
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document.write( "So, the second equation is\r\n" );
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document.write( "    x^2 - qx + 9 = 0.\r\n" );
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document.write( "Since it has equal roots,  these roots are EITHER\r\n" );
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document.write( "    a)  both are equal to   =  3   (Vieta;s theorem).\r\n" );
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document.write( "OR\r\n" );
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document.write( "    b)  both are equal to   = -3   (Vieta;s theorem).\r\n" );
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document.write( "In case a),  q= 3 + 3 = 6.   (Vieta;s theorem, again)    ANSWER\r\n" );
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document.write( "In case b),  q = (-3) + (-3) = -6.\r\n" );
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document.write( "So, the problem has TWO answers:  q = 6  and/or  q = -6.\r\n" );
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\n" ); document.write( "\n" ); document.write( "Again : one possible answer is q = 6; the other possible answer is q = -6.\r
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\n" ); document.write( "\n" ); document.write( "This problem is to apply Vieta's theorem 4 times.\r
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