What is (x(x+2))/(4x^2-9)< 1/5 ?

The solution to (x(x+2))/(4x^2-9)< 1/5 is -9<x<-3/2 \lor-1<x< 3/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x(x+2))/(4x^2-9)< 1/5

\frac{x ( x + 2 )}{4 x ^{2} - 9} < \frac{1}{5}

- 9 < x < - \frac{3}{2} or - 1 < x < \frac{3}{2}

Interval Notation

(- 9, - \frac{3}{2}) \cup (- 1, \frac{3}{2})

Decimal Notation

- 9 < x < - 1.5 or - 1 < x < 1.5

Solution steps

\frac{x ( x + 2 )}{4 x ^{2} - 9} < \frac{1}{5}

Rewrite in standard form

\frac{x ^{2} + 10 x + 9}{( 2 x + 3 ) ( 2 x - 3 )} < 0

Factor

\frac{x ^{2} + 10 x + 9}{( 2 x + 3 ) ( 2 x - 3 )} : \frac{( x + 1 ) ( x + 9 )}{( 2 x + 3 ) ( 2 x - 3 )}

\frac{( x + 1 ) ( x + 9 )}{( 2 x + 3 ) ( 2 x - 3 )} < 0

Identify the intervals

- 9 < x < - \frac{3}{2} or - 1 < x < \frac{3}{2}

64 x^{3} - y^{6}

\frac{q ^{6} ( - r ) ^{7}}{q ^{10} r ^{5}}

2^{l o g_{2} (6)}

100 f - 75 = 50 f + 75

- 5 < x - 4 < 2 - x

What is (x(x+2))/(4x^2-9)< 1/5 ?
The solution to (x(x+2))/(4x^2-9)< 1/5 is -9<x<-3/2 \lor-1<x< 3/2