Question

Solve the inequality x^2+2x>=0

Answer 1

\[x(x + 2) \ge 0\]

Answer 2

Observe that if you want \(x \times (x + 2)\) nonnegative, both \(x\) and \(x + 2\) are nonnegative.

Answer 3

yes

Answer 4

hm

Answer 5

First time I am solving a 2nd degree inequality :S

Answer 6

Could someone help me continue? :)

Answer 7

I can offer an alternative. Just sketch the graph. That never fails.

Answer 8

I need to do it without a graph

Answer 9

\[x \le -2 \] and \[x \ge0\]

Answer 10

How and why?

Answer 11

Okay, then first find the x-axis intercepts by treating the inequality sign as an equals sign.
\[x ^{2}+2x=0\]
\[x(x+2)=0\]
\[x=-2,0\]
Sub in values for \[x<-2\]
e.g when x=-3, y=3

Then sub in values for x>0

e.g when x=1, y=3

Since \[x \in (-\infty,-2)\]and \[x \in (0,\infty)\]yield positive answers

Therefore \[x ^{2}+2x>=0\]
for x=<-2 and x>=0

Answer 12

Two factors: x and x+2
If the product is positive, x and x+2 must both be positive OR x and x+2 must both be negative.

Let us make a picture of the factors and their signs at various points on the number line: