|x^2-1|=|2x-1|

解

x^{2} - 1 = ∣ 2 x - 1 ∣

x = - 1 - 3 or x = 0 or x = 3 - 1 or x = 2

十進法表記

x = - 2.73205 \dots, x = 0, x = 0.73205 \dots, x = 2

解答ステップ

x^{2} - 1 = ∣ 2 x - 1 ∣

正と負の区間を求める

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

各区間の不等式を解く

x = - 1 - 3 or x = 0 or x = 3 - 1 or x = 2