What is 1/((x^2+x))<= 1/((2x^2+2x+3)) ?

The solution to 1/((x^2+x))<= 1/((2x^2+2x+3)) is -1<x<0

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

1/((x^2+x))<= 1/((2x^2+2x+3))

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

- 1 < x < 0

Interval Notation

(- 1, 0)

Solution steps

\frac{1}{x ^{2} + x} \leq \frac{1}{2 x ^{2} + 2 x + 3}

Rewrite in standard form

\frac{x ^{2} + x + 3}{x ( x + 1 ) ( 2 x ^{2} + 2 x + 3 )} \leq 0

Identify the intervals

- 1 < x < 0

\frac{2}{x - 1} \leq \frac{3}{2 x - 3}

(x - 3) (x - 7) (x - 12) < 0

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

\frac{( x - 5 )}{( x ^{2} + 5 x - 14 )} > 0

\frac{1}{( 3 x - 8 )} \leq \frac{1}{( 8 x - 3 )}

What is 1/((x^2+x))<= 1/((2x^2+2x+3)) ?
The solution to 1/((x^2+x))<= 1/((2x^2+2x+3)) is -1<x<0