What is (2x)/(5-x^2)<= 1/(8x) ?

The solution to (2x)/(5-x^2) sqrt(5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(2x)/(5-x^2)<= 1/(8x)

\frac{2 x}{5 - x ^{2}} \leq \frac{1}{8 x}

- 5 < x \leq - \frac{5}{17} or 0 < x \leq \frac{5}{17} or x > 5

Interval Notation

(- 5, - \frac{5}{17}] \cup (0, \frac{5}{17}] \cup (5, \infty)

Decimal Notation

- 2.23606 \dots < x \leq - 0.54232 \dots or 0 < x \leq 0.54232 \dots or x > 2.23606 \dots

Solution steps

\frac{2 x}{5 - x ^{2}} \leq \frac{1}{8 x}

Rewrite in standard form

\frac{- 17 x ^{2} + 5}{x ( x + 5 ) ( x - 5 )} \leq 0

Factor

\frac{- 17 x ^{2} + 5}{x ( x + 5 ) ( x - 5 )} : \frac{- ( 17 x + 5 ) ( 17 x - 5 )}{x ( x + 5 ) ( x - 5 )}

\frac{- ( 17 x + 5 ) ( 17 x - 5 )}{x ( x + 5 ) ( x - 5 )} \leq 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - ( 17 x + 5 ) ( 17 x - 5 ) ) ( - 1 )}{x ( x + 5 ) ( x - 5 )} \geq 0 \cdot (- 1)

Simplify

\frac{( 17 x + 5 ) ( 17 x - 5 )}{x ( x + 5 ) ( x - 5 )} \geq 0

Identify the intervals

- 5 < x \leq - \frac{5}{17} or 0 < x \leq \frac{5}{17} or x > 5

- 36 > 6 x + 12

- 5 + (\frac{1}{2}) x \leq 3

- 68

\frac{1}{8} x = 5

x^{2} + 11 x + 28 = 0

What is (2x)/(5-x^2)<= 1/(8x) ?
The solution to (2x)/(5-x^2)<= 1/(8x) is -sqrt(5)<x<=-sqrt(5/17)\lor 0<x<= sqrt(5/17)\lor x>sqrt(5)