How do you solve #x- 4> \frac { 7} { x }#?

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Answer 1 · 2017-01-27T16:03:14

#x < 2-sqrt11 and >2+sqrt11#.

#x ne 0#.

When #x > 0, x(x-4) > 7#, giving

(x-2)^2-4 > 11, and so,

#x > 2 +sqrt11#.

When #x < 0, x(x-4),> 7#, giving

(x-2)^2-4 < 11, and so,

#x < 2 -sqrt11#.

Thus, #x in R, sans [2-sqrt11, 2+sqrt11]#.