What is |(x^2-2x-5)/(x^2+4x-7)|<1 ?

The solution to |(x^2-2x-5)/(x^2+4x-7)| 2

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|(x^2-2x-5)/(x^2+4x-7)|<1

\frac{x ^{2} - 2 x - 5}{x ^{2} + 4 x - 7} < 1

- 3 < x < \frac{1}{3} or x > 2

Interval Notation

(- 3, \frac{1}{3}) \cup (2, \infty)

Decimal Notation

- 3 < x < 0.33333 \dots or x > 2

Solution steps

\frac{x ^{2} - 2 x - 5}{x ^{2} + 4 x - 7} < 1

Apply absolute rule: If

∣ u ∣ < a, a > 0

then

- a < u < a

- 1 < \frac{x ^{2} - 2 x - 5}{x ^{2} + 4 x - 7} < 1

\frac{x ^{2} - 2 x - 5}{x ^{2} + 4 x - 7} > - 1 and \frac{x ^{2} - 2 x - 5}{x ^{2} + 4 x - 7} < 1

(x < - 2 - 11 or - 3 < x < - 2 + 11 or x > 2) and (- 2 - 11 < x < \frac{1}{3} or x > - 2 + 11)

Merge Overlapping Intervals

- 3 < x < \frac{1}{3} or x > 2

c^{2} + 6 c + 9

4 (x - 7)^{2} + 20 = 24

5 (- 2 - 3 n) = - 33 + 8 n

- 10 - 7

24 = \frac{6}{7} (k + 8)

What is |(x^2-2x-5)/(x^2+4x-7)|<1 ?
The solution to |(x^2-2x-5)/(x^2+4x-7)|<1 is -3<x< 1/3 \lor x>2