What is |((x^2-3x-1))/((x^2+x+1))|<3 ?

The solution to |((x^2-3x-1))/((x^2+x+1))| -1

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|((x^2-3x-1))/((x^2+x+1))|<3

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

x < - 2 or x > - 1

Interval Notation

(- \infty, - 2) \cup (- 1, \infty)

Solution steps

\frac{x ^{2} - 3 x - 1}{x ^{2} + x + 1} < 3

Apply absolute rule: If

∣ u ∣ < a, a > 0

then

- a < u < a

- 3 < \frac{x ^{2} - 3 x - 1}{x ^{2} + x + 1} < 3

\frac{x ^{2} - 3 x - 1}{x ^{2} + x + 1} > - 3 and \frac{x ^{2} - 3 x - 1}{x ^{2} + x + 1} < 3

True for all x and (x < - 2 or x > - 1)

Merge Overlapping Intervals

x < - 2 or x > - 1

∣ x ∣ < \frac{2}{x}

x^{2} - ∣ x ∣ - 2 \geq 0

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

\frac{1}{( ∣ x ∣ - 3 )} < \frac{1}{2}

3 ∣ x - 1 ∣ + x^{2} - 7 > 0

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