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

The solution to |x^2+3x-1|<3 is -4<x<-2\lor-1<x<1

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|x^2+3x-1|<3

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

- 4 < x < - 2 or - 1 < x < 1

Interval Notation

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

Solution steps

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

Apply absolute rule: If

∣ u ∣ < a, a > 0

then

- a < u < a

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

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

(x < - 2 or x > - 1) and - 4 < x < 1

Merge Overlapping Intervals

- 4 < x < - 2 or - 1 < x < 1

0.0657 - (0.7 5^{2} \times 0.0676)

2 x^{2} = 100

\frac{6 ^{- 5}}{6 ^{3}}

\frac{3 ^{2}}{3}

- 1 \leq \frac{x}{2 x - 3} \leq 1