High School Algebra

[ECHOSIDE]

Hello! I am taking a high school algebra course at my local university and am having trouble with a certain type of quadratic equation. I figured someone in these forums might be able to help. It's probably something simple that I'm not understanding.
I actually don't have trouble solving the equation, it's more a question of how the equation works.
Starting with the equation [x+(1/x)=(13/6)], I perform the following steps to put it into the standard quadratic form:
x+(1/x)=(13/6)
(1/x)=-x+(13/6)
1=-x^2+(13x/6)
0=-x^2+(13x/6)-1

The trouble I'm having is that [x+(1/x)=(13/6)] and [0=-x^2+(13x/6)-1] produce different graphs with different x-intercepts.
If the equations are equivalent, why do they produce different x-intercepts?
If the equations are not equivalent, how can manipulating them in this way be mathematically justifiable?

Thanks

 

[ECHOSIDE]

I figured it out (with the help of a tutor)!
It turns out that while the equations do produce different graphs, the x-intercepts are the same. On my graphing calculator, I had not zoomed in far enough to see that x+(1/x)=(13/6) dips below the x-axis in quadrant IV - producing 2 x-intercepts at (1/2, 0) and at (3/2, 0). These are the same intercepts produced for 0=-x^2+(13x/6)-1.

 

[Sequences]

You need to be careful with this equation. If 0 is in the domain, then multiplying by x is not recommended.