What is (3x+3)/(x^2-5x-4)>= 0 ?

The solution to (3x+3)/(x^2-5x-4)>= 0 is -1 (5+sqrt(41))/2

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(3x+3)/(x^2-5x-4)>= 0

\frac{3 x + 3}{x ^{2} - 5 x - 4} \geq 0

- 1 \leq x < \frac{5 - 41}{2} or x > \frac{5 + 41}{2}

Interval Notation

[- 1, \frac{5 - 41}{2}) \cup (\frac{5 + 41}{2}, \infty)

Decimal Notation

- 1 \leq x < - 0.70156 \dots or x > 5.70156 \dots

Solution steps

\frac{3 x + 3}{x ^{2} - 5 x - 4} \geq 0

Factor

\frac{3 x + 3}{x ^{2} - 5 x - 4} : \frac{3 ( x + 1 )}{( x - \frac{5 + 41}{2} ) ( x - \frac{5 - 41}{2} )}

\frac{3 ( x + 1 )}{( x - \frac{5 + 41}{2} ) ( x - \frac{5 - 41}{2} )} \geq 0

Divide both sides by

3

\frac{\frac{3 ( x + 1 )}{( x - \frac{5 + 41}{2} ) ( x - \frac{5 - 41}{2} )}}{3} \geq \frac{0}{3}

Simplify

\frac{x + 1}{( x - \frac{5 + 41}{2} ) ( x - \frac{5 - 41}{2} )} \geq 0

Identify the intervals

- 1 \leq x < \frac{5 - 41}{2} or x > \frac{5 + 41}{2}

\frac{- 3 x}{4} < 3 or 4 x + 6 < - 3

- 2.3 (x - 1.2) = - 9.66

\frac{120}{2 π} (\frac{1}{2} - \frac{1}{18} - \frac{1}{4} - \frac{1}{14})

22 \geq 5 (2 y + 3) - 3 y

What is (3x+3)/(x^2-5x-4)>= 0 ?
The solution to (3x+3)/(x^2-5x-4)>= 0 is -1<= x<(5-sqrt(41))/2 \lor x>(5+sqrt(41))/2