What is | x/((x^2-x-1))|<= 2 ?

The solution to | x/((x^2-x-1))| = 2

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| x/((x^2-x-1))|<= 2

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

x \leq \frac{1 - 17}{4} or - \frac{1}{2} \leq x \leq \frac{1 + 17}{4} or x \geq 2

Interval Notation

(- \infty, \frac{1 - 17}{4}] \cup [- \frac{1}{2}, \frac{1 + 17}{4}] \cup [2, \infty)

Decimal Notation

x \leq - 0.78077 \dots or - 0.5 \leq x \leq 1.28077 \dots or x \geq 2

Solution steps

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

Apply absolute rule: If

∣ u ∣ \leq a, a > 0

then

- a \leq u \leq a

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

\frac{x}{x ^{2} - x - 1} \geq - 2 and \frac{x}{x ^{2} - x - 1} \leq 2

(x \leq \frac{1 - 17}{4} or \frac{1 - 5}{2} < x \leq \frac{1 + 17}{4} or x > \frac{1 + 5}{2}) and (x < \frac{1 - 5}{2} or - \frac{1}{2} \leq x < \frac{1 + 5}{2} or x \geq 2)

Merge Overlapping Intervals

x \leq \frac{1 - 17}{4} or - \frac{1}{2} \leq x \leq \frac{1 + 17}{4} or x \geq 2

∣ 3 x - 4 ∣ > 2 - 5 x

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

x^{2} - 6 x + 9 \geq 4

\frac{3}{x + 1} > 2

∣ 2 x ∣ 3 - x ∣ - 2 ∣ \leq 4

What is | x/((x^2-x-1))|<= 2 ?
The solution to | x/((x^2-x-1))|<= 2 is x<= (1-sqrt(17))/4 \lor-1/2 <= x<= (1+sqrt(17))/4 \lor x>= 2