What is ((1-2x-3x^2))/((3x-x^2-5))>= 3 ?

The solution to ((1-2x-3x^2))/((3x-x^2-5))>= 3 is x>= 16/11

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((1-2x-3x^2))/((3x-x^2-5))>= 3

\frac{( 1 - 2 x - 3 x ^{2} )}{( 3 x - x ^{2} - 5 )} \geq 3

x \geq \frac{16}{11}

Interval Notation

[\frac{16}{11}, \infty)

Decimal Notation

x \geq 1.45454 \dots

Solution steps

\frac{1 - 2 x - 3 x ^{2}}{3 x - x ^{2} - 5} \geq 3

Rewrite in standard form

\frac{- 11 x + 16}{3 x - x ^{2} - 5} \geq 0

Multiply both sides by

- 1

(reverse the inequality)

\frac{( - 11 x + 16 ) ( - 1 )}{- ( - 3 x + x ^{2} + 5 )} \leq 0 \cdot (- 1)

Simplify

\frac{- 11 x + 16}{- 3 x + x ^{2} + 5} \leq 0

Identify the intervals

x \geq \frac{16}{11}

x^{4} + x^{3} - x - 1 > 0

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

x^{4} - 2 x^{3} - x + 2 > 0

x^{2} - 3 x + 5 > 0

(x + 4)^{5} \cdot (x + 3)^{6} \cdot (x + 2)^{7} \cdot (x - 1)^{8} \leq 0

What is ((1-2x-3x^2))/((3x-x^2-5))>= 3 ?
The solution to ((1-2x-3x^2))/((3x-x^2-5))>= 3 is x>= 16/11