How do you solve # x/2>1+(4/x)#?

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Answer 1 · 2016-05-29T07:18:13

x is outside #[4, -2]#

x cannot be 0.

Multiply throughout by 2x and rearrange. The inequality sign is

reversed, when x is negative. Accordingly,

#x^2-2x-8 > < 0, respectively.

#(x-1)^2 > < 9#. So, x-1 > < +- 3>

In the first case, x > 0, #and so, x > 4; the other inequality is

inadmissible.

When x is negative, #x < -2#; the other inequality is inadmissible.

So, #x < -2 and x > 4#