What is (x^2+4x-1)/((x^2+1)(x+2))>0 ?

The solution to (x^2+4x-1)/((x^2+1)(x+2))>0 is -2-sqrt(5) -2+sqrt(5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x^2+4x-1)/((x^2+1)(x+2))>0

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

- 2 - 5 < x < - 2 or x > - 2 + 5

Interval Notation

(- 2 - 5, - 2) \cup (- 2 + 5, \infty)

Decimal Notation

- 4.23606 \dots < x < - 2 or x > 0.23606 \dots

Solution steps

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

Factor

x^{2} + 4 x - 1 : (x + 2 - 5) (x + 2 + 5)

\frac{( x + 2 - 5 ) ( x + 2 + 5 )}{( x ^{2} + 1 ) ( x + 2 )} > 0

Identify the intervals

- 2 - 5 < x < - 2 or x > - 2 + 5

3 x + 9 = 1 - x

0 + \frac{π}{12}

x - 1 \leq - 9

x + 9 = 16

∣ x - 2 ∣ = - 9

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