# Thread: urgent . tmorrow sch

1. ## urgent . tmorrow sch

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)

2. Hello,

This is because
(x+1)^2 > 0
is always TRUE.

Unless (x+1)²=0, because it's $>$, not $\geq$

And (x+1)²=0 if and only if x=... ?

3. Originally Posted by Moo
Hello,

This is because

is always TRUE.

Unless (x+1)²=0, because it's $>$, not $\geq$

And (x+1)²=0 if and only if x=... ?

ok thx for clarification.

here is another question.

(x-2)^2 (x+1) less than or equal 0

how to find the range of values of x

x less than or equal to -1, while x=2 ( why)??

4. $(x-2)^2 (x+1) \le 0$

When $x=2$ then $(x-2)^2 = 0$ so $(x-2)^2 (x+1) = 0$ so that gives you the $x=2$ solution.

Otherwise $(x-2)^2 > 0$ and so $(x-2)^2 (x+1) \le 0$ whenever $(x+1) \le 0$ or when $x \le -1$.

Put them together:

$(x-2)^2 (x+1) \le 0 \Longrightarrow x=2 \textrm{ or } x \le -1$.

So it's:
"x less than or equal to -1, or x=2" not "x less than or equal to -1, while x=2".