What is 1>= 3/((3-2x-x^2)) ?

The solution to 1>= 3/((3-2x-x^2)) is x 1

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1>= 3/((3-2x-x^2))

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

x < - 3 or - 2 \leq x \leq 0 or x > 1

Interval Notation

(- \infty, - 3) \cup [- 2, 0] \cup (1, \infty)

Solution steps

1 \geq \frac{3}{3 - 2 x - x ^{2}}

Switch sides

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

Rewrite in standard form

\frac{x ^{2} + 2 x}{3 - 2 x - x ^{2}} \leq 0

Factor

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

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

Multiply both sides by

- 1

(reverse the inequality)

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

Simplify

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

Identify the intervals

x < - 3 or - 2 \leq x \leq 0 or x > 1

x^{2} + 3 x + 1 < 0

121 x^{2} - 22 x + 1 < 0

(x^{2} + 3 x + 13) > 0

(x + 1)^{2} \cdot (x - 1)^{3} \cdot (x - 5)^{7} \geq 0

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

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