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

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