find the range of values of x for each of the following inequalities
x^2 + 2x +1 >0
(x+1)^2 > 0
(x+1) >0
x>1
how come the answer given is ( x is a subset of IR, x is not equal to -1)
$\displaystyle (x-2)^2 (x+1) \le 0$
When $\displaystyle x=2$ then $\displaystyle (x-2)^2 = 0$ so $\displaystyle (x-2)^2 (x+1) = 0$ so that gives you the $\displaystyle x=2$ solution.
Otherwise $\displaystyle (x-2)^2 > 0$ and so $\displaystyle (x-2)^2 (x+1) \le 0$ whenever $\displaystyle (x+1) \le 0$ or when $\displaystyle x \le -1$.
Put them together:
$\displaystyle (x-2)^2 (x+1) \le 0 \Longrightarrow x=2 \textrm{ or } x \le -1$.
So it's:
"x less than or equal to -1, or x=2" not "x less than or equal to -1, while x=2".