Question 750840
<pre>
 ax²+bx+c=0

The solutions are

{{{(-b + sqrt( b^2-4*a*c ))/(2*a) }}} and {{{(-b - sqrt( b^2-4*a*c ))/(2*c) }}} 


 cy²+by+a=0

The solutions are

{{{(-b + sqrt( b^2-4*a*c ))/(2*c) }}} and {{{(-b - sqrt( b^2-4*a*c ))/(2*c) }}}

They are reciprocals then their product must be 1:

Let's multiply the solution of the first with a positive radical times
the solution of the second with a negative radical:

{{{((-b + sqrt( b^2-4*a*c ))/(2*a)) }}}{{{""*""}}}{{{((-b - sqrt( b^2-4*a*c ))/(2*c) ))}}} 



{{{(-b + sqrt( b^2-4*a*c ))(-b - sqrt( b^2-4*a*c ))/(4*a*c) }}}

The two factors in the numerators are conjugates so FOILing them
causes outers and inners to cancel, so we get:

{{{((-b)^2 - (sqrt(b^2-4ac))^2)/(4ac) }}} =

{{{(b^2-(b^2-4ac))/(4ac)}}} =

{{{(b^2-b^2+4ac)/(4ac)}}} =

{{{(4ac)/(4ac)}}} =

1

so they are reciprocals.  Multiplying the other pair
gives the same results.

Edwin</pre>