$\frac { 2 x } { x + 1 } \leq \frac { 4 } { x - 5 }$
A) $- 1

Question

$\frac { 2 x } { x + 1 } \leq \frac { 4 } { x - 5 }$
A) $- 1 < x \leq \frac { 7 - \sqrt { 55 } } { 2 } \text { or } 5 < x \leq \frac { 7 + \sqrt { 55 } } { 2 }$
B) $- 1 < x \leq \frac { 7 + \sqrt { 57 } } { 2 } \text { or } 5 < x \leq \frac { 7 - \sqrt { 57 } } { 2 }$
C) $- 1 < x < \frac { 7 - \sqrt { 55 } } { 2 } \text { or } 5 < x < \frac { 7 + \sqrt { 57 } } { 2 }$
D) $- 1 < x \leq \frac { 7 - \sqrt { 57 } } { 2 } \text { or } 5 < x \leq \frac { 7 + \sqrt { 57 } } { 2 }$

Quizplus · Accepted Answer

To solve the inequality $\frac { 2 x } { x + 1 } \leq \frac { 4 } { x - 5 }$, we first find a common denominator and bring all terms to one side to form a quadratic inequality. Solving this inequality involves finding the roots of the corresponding quadratic equation, which, after simplification, yields $x = \frac { 7 \pm \sqrt { 57 } } { 2 }$. The solution set must exclude the values that make the original denominators zero, which are $x = -1$ and $x = 5$, leading to the correct intervals as given in option D.

2xx+1≤4x−5\frac { 2 x } { x + 1 } \leq \frac { 4 } { x - 5 }x+12x​≤x−54​

$\frac { 2 x } { x + 1 } \leq \frac { 4 } { x - 5 }$