Repost! :)

Not totally sure about this one.....but.....here's what I found

Using the  second equation

ax^2   +  bx  +  1   =  0

ax^2  +  bx  =  -1

x ( ax + b)  =   -1

ax + b =    -1/x

Sub this into the first equation

x^2  -  1/x   =    0      multiply through  by  x

x^3   -  1   =    0     factor as a difference  of  cubes

( x - 1)  (x^2  + x + 1)  =  0

Only the first   produces a  single real solution     x   = 1

So....using the first equation 

(1)^2  + a(1)  +  b   = 0

1  +  a +  b  =   0

a  +  b   =  -1

Hi Cphill, u are completely corrrect, but they ask for multiple values, I suspect in $(x-1)(x^2+x+1)$, we also have to solve for $(x^2+x+1)=0$ to get the other 2 values since it always have two values.

I'm guessing because $r^2+r+1=0$

   comparing:              $ax^2+bx+1=0$

therefore a=1, b=1, therefore resulting in $a+b=2$