What is ((x^2+4))/((5x))>= 1x ?

The solution to ((x^2+4))/((5x))>= 1x is x<=-1\lor 0<x<= 1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

((x^2+4))/((5x))>= 1x

\frac{( x ^{2} + 4 )}{( 5 x )} \geq 1 x

x \leq - 1 or 0 < x \leq 1

Interval Notation

(- \infty, - 1] \cup (0, 1]

Solution steps

\frac{x ^{2} + 4}{5 x} \geq 1 \cdot x

Rewrite in standard form

\frac{- 4 x ^{2} + 4}{x} \geq 0

Factor

\frac{- 4 x ^{2} + 4}{x} : \frac{- 4 ( x + 1 ) ( x - 1 )}{x}

\frac{- 4 ( x + 1 ) ( x - 1 )}{x} \geq 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - 4 ( x + 1 ) ( x - 1 ) ) ( - 1 )}{x} \leq 0 \cdot (- 1)

Simplify

\frac{4 ( x + 1 ) ( x - 1 )}{x} \leq 0

Divide both sides by

4

\frac{\frac{4 ( x + 1 ) ( x - 1 )}{x}}{4} \leq \frac{0}{4}

Simplify

\frac{( x + 1 ) ( x - 1 )}{x} \leq 0

Identify the intervals

x \leq - 1 or 0 < x \leq 1

1 - x^{3} < 0

2 > \frac{( 6 x + 28 )}{( x + 2 )}

\frac{( e ) ( x ^{2} - 7 x + 12 )}{( 2 x ^{2} + 4 x + 5 )} > \frac{1}{2}

x^{2} \cdot (x - 3) < 0

\frac{( x ^{3} + x ^{2} + x + 1 ) ( x - 1 )}{( x ^{4} - 4 )} \geq 0

What is ((x^2+4))/((5x))>= 1x ?
The solution to ((x^2+4))/((5x))>= 1x is x<=-1\lor 0<x<= 1