Question

find the values of x for which x^2-3x+2/(x+1) is euqal to or greater than 0.

Answer 1

but the answe is -1<x<1 and x>2. how should i do?

Answer 2

\[-1<x \le1, x \ge2\]

Answer 3

\[\frac{x^2-3x+2}{x+1}\geq 0\]\[(x^2-3x+2\geq0\quad\cap\quad x+1>0)\cup(x^2-3x+2\leq0\quad\cap\quad x+1<0)\]\[(x\in(-\infty,1]\cup[2,+\infty)\quad\cap\quad x>-1)\cup(1\leq x\leq2\quad\cap\quad x<-1)\]\[x\in(-1,1]\cup[2,+\infty)\quad\cup\quad\emptyset\]\[\Rightarrow\quad x\in(-1,1]\cup[2,+\infty)\]

Answer 4

i have written the same answer !