Given $x^2-3x-1=0,$ what is $\dfrac{1}{x}-x ?$
(please no decimal answers)
Thanks for helping!
Add 1 to both sides: $x^2-3x=1$
Divide by x: $x-3 = \frac{1}{x}$
Subtract x from both sides: $-3 = \frac{1}{x} - x$
Our answer is $$-3$$
+421 
