What is |x^2-2x-3|>|x^2+7x-13| ?

The solution to |x^2-2x-3|>|x^2+7x-13| is x<(-3sqrt(17)-5)/4 \lor 10/9 <x<(3sqrt(17)-5)/4

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|x^2-2x-3|>|x^2+7x-13|

x^{2} - 2 x - 3 > x^{2} + 7 x - 13

x < \frac{- 3 17 - 5}{4} or \frac{10}{9} < x < \frac{3 17 - 5}{4}

Interval Notation

(- \infty, \frac{- 3 17 - 5}{4}) \cup (\frac{10}{9}, \frac{3 17 - 5}{4})

Decimal Notation

x < - 4.34232 \dots or 1.11111 \dots < x < 1.84232 \dots

Solution steps

x^{2} - 2 x - 3 > x^{2} + 7 x - 13

Find positive and negative intervals

x \leq \frac{- 101 - 7}{2}, \frac{- 101 - 7}{2} < x \leq - 1, - 1 < x < \frac{101 - 7}{2}, \frac{101 - 7}{2} \leq x < 3, x \geq 3

Solve the inequality for each interval

x \leq \frac{- 101 - 7}{2} or \frac{- 101 - 7}{2} < x < \frac{- 3 17 - 5}{4} or \frac{10}{9} < x < \frac{101 - 7}{2} or \frac{101 - 7}{2} \leq x < \frac{3 17 - 5}{4} or No Solution

Merge Overlapping Intervals

x < \frac{- 3 17 - 5}{4} or \frac{10}{9} < x < \frac{3 17 - 5}{4}

\frac{( 2 x ^{2} - 4 )}{3} \geq 2 x^{2}

∣ x + 1 ∣ < ∣ 2 x - 1 ∣

\frac{( 5 x + 5 )}{10} \geq 5

x^{2} - 2 x - 3 > 1

∣ x - 2 ∣ + ∣ x - 3 ∣ + ∣ x - 4 ∣ > 6

What is |x^2-2x-3|>|x^2+7x-13| ?
The solution to |x^2-2x-3|>|x^2+7x-13| is x<(-3sqrt(17)-5)/4 \lor 10/9 <x<(3sqrt(17)-5)/4