Question 222372
<pre><font size = 4 color = "indigo"><b>
This has a "surprise" answer:

{{{6x^2+kx-6=0}}}

The solutions by the quadratic equation are:

{{{x = (-k +- sqrt( k^2-4*(6)*-6) )/(2*6) }}} 

{{{x = (-k +- sqrt( k^2-4*(6)*-6) )/(2*6) }}} 

{{{x = (-k +- sqrt( k^2+144) )/12 }}} 

So there are two solutions:

(-k + sqrt( k^2+144) )/12 and (-k - sqrt( k^2+144) )/12

Any number times its negative reciprocal must equal -1.

So we multiply these two roots and set the product 
equal to -1:

{{{( (-k + sqrt( k^2+144) )/12) ((-k - sqrt( k^2+144) )/12)=-1}}}

Multiply using "FOIL"

{{{( (k^2 - ( k^2+144) )/144)=-1}}}

{{{( (k^2 -  k^2-144 )/144)=-1}}}

{{{ -144/144=-1}}}

{{{-1=-1}}}

Since this comes out an identity, that means that
{{{k}}} can be any number whatsoever, and in every
case one root will be the negative reciprocal of
the other!

Edwin</pre>