What is (4x^2-25)/(x^2-9)<= 0 ?

The solution to (4x^2-25)/(x^2-9)<= 0 is -3<x<=-5/2 \lor 5/2 <= x<3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(4x^2-25)/(x^2-9)<= 0

\frac{4 x ^{2} - 25}{x ^{2} - 9} \leq 0

- 3 < x \leq - \frac{5}{2} or \frac{5}{2} \leq x < 3

Interval Notation

(- 3, - \frac{5}{2}] \cup [\frac{5}{2}, 3)

Decimal Notation

- 3 < x \leq - 2.5 or 2.5 \leq x < 3

Solution steps

\frac{4 x ^{2} - 25}{x ^{2} - 9} \leq 0

Factor

\frac{4 x ^{2} - 25}{x ^{2} - 9} : \frac{( 2 x + 5 ) ( 2 x - 5 )}{( x + 3 ) ( x - 3 )}

\frac{( 2 x + 5 ) ( 2 x - 5 )}{( x + 3 ) ( x - 3 )} \leq 0

Identify the intervals

- 3 < x \leq - \frac{5}{2} or \frac{5}{2} \leq x < 3

- 13 x = 26

- 5 \leq \frac{2 - 3 x}{6} \leq 2

216 pq - p^{7} q

\frac{3}{9} = \frac{w}{6}

3 x^{4} + 6 x^{3} + 9 x^{2}

What is (4x^2-25)/(x^2-9)<= 0 ?
The solution to (4x^2-25)/(x^2-9)<= 0 is -3<x<=-5/2 \lor 5/2 <= x<3