Solving inequality

**metlx** · December 14th 2009, 12:09 PM

See this example:

$\frac{x - 1}{x + 1} \geq \frac{2x}{x - 1}$

We multiply both sides by $(x + 1) (x - 1)$

and get:

$(x - 1)^2 \geq 2x(x +1)$

Criteria 1:

x < -1
$(x - 1)^2 \geq 2x(x +1)$
$x^2 - 2x + 1 \geq 2x^2 + 2x$
$x^2 + 4x -1 \leq 0$

$x_{1,2} = \frac{-4 \pm 2\sqrt{5}}{2} = -2 \pm \sqrt{5}$

$x \in [-2-\sqrt{5}, -2+\sqrt{5}]$ but since $x < -1$ we get $x \in [-2-\sqrt{5}, 1)$

Criteria 2:
$x \in (-1, 1)$

$(x - 1)^2\leq 2x(x +1)$ why less or equal??

from that we get: $x^2 +4x -1 \geq 0 \Rightarrow x \in [-2+\sqrt{5},1)$ (why not (-1, -2+\sqrt{5}] ?)

Criteria 3:
x > 1
$x^2 + 4x - 1 \leq 0 \Rightarrow no solution$

All together:

$x \in [-2-\sqrt{5},-1) U [-2+\sqrt{5},1)$

**NOX Andrew** · December 14th 2009, 01:17 PM

If $x \in (-1,1)$ , then $(x - 1)(x + 1)$ is negative.

**metlx** · December 14th 2009, 01:51 PM

short and simple. thanks

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