Question

(1/(x+1)) <= -(1/2)

Answer 1

\[\frac{1}{x+1} \le -\frac{1}{2}\]

I keep getting \[0 \ge (x+3)(x+1)\]

WolframAlpha says it's \[(x+2)(x+3)\]

Answer 2

This is what I have so far:

First multiply both sides by square of the denominator to get
\[x+1 \le -\frac{1}{2}(x+1)^2\]


Then multiply by -2 to clear fraction
\[-2(x+1) \ge (x+1)^2\]

Then I factorise and get
\[0 \ge (x+3)(x+1)\]

Answer 3

so do u wanna know the value of x

Answer 4

I'm trying to solve for x.

Answer 5

it's too easy
\[0/(x+1) \ge (x+3)\]\[0 \ge (x+3)\]\[-3 \ge x\]

Answer 6

\[0/(x+3)\ge(x+1)\]\[-1 \ge x\]

Answer 7

gt it @thrustfault ?????

Answer 8

Thanks.

Answer 9

yw:)