document.write( "Question 1086139: Suppose a and x satisfy x^2 + (a-(1/a))x - 1 = 0. Solve for x in terms of a.\r
\n" ); document.write( "\n" ); document.write( "*That's a quadratic in x \n" ); document.write( "

Algebra.Com's Answer #700348 by ikleyn(52787) $\"\"$ $\"About$

You can put this solution on YOUR website!
.
\n" ); document.write( "I walked in the park and thought on this problem.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "I got two solutions. Each is as short as 2 - 4 lines.\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Solution 1 (the Vieta's theorem; 4-lines solution)\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "

\r\n" );
document.write( "The Vieta's theorem says:  if  p and q are the roots of a quadratic equation   = 0  then  u = -(p+q)  and  v = pq.\r\n" );
document.write( "\r\n" );
document.write( "The opposite is also TRUE:  if  u = -(p+q)  and  v = pq  then p and q  are the roots of the quadratic equation     = 0.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Now look into your equation and notice that  the numbers  -a  and    give   when summed up and -1 when multiplied.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Hence,  the numbers \"-a\" and   are the roots of your equation.\r\n" );
document.write( "

\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Solution 2 (Factoring. 2-lines solution)\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "

\r\n" );
document.write( "Factor your polynomial:   = .\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Hence, the roots are \"-a\" and .\r\n" );
document.write( "

\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

\n" );