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

Jan 27, 2017

$x < 2 - \sqrt{11} \mathmr{and} > 2 + \sqrt{11}$.

#### Explanation:

$x \ne 0$.

When $x > 0 , x \left(x - 4\right) > 7$, giving

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

$x > 2 + \sqrt{11}$.

When $x < 0 , x \left(x - 4\right) , > 7$, giving

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

$x < 2 - \sqrt{11}$.

Thus, $x \in R , s a n s \left[2 - \sqrt{11} , 2 + \sqrt{11}\right]$.
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