|x^2-6x+1|=|(3x+5)/2 |

解

x^{2} - 6 x + 1 = \frac{3 x + 5}{2}

x = \frac{15 - 249}{4} or x = 1 or x = \frac{7}{2} or x = \frac{15 + 249}{4}

十進法表記

x = - 0.19493 \dots, x = 1, x = 3.5, x = 7.69493 \dots

解答ステップ

x^{2} - 6 x + 1 = \frac{3 x + 5}{2}

正と負の区間を求める

x < - \frac{5}{3}, - \frac{5}{3} \leq x \leq - 22 + 3, - 22 + 3 < x < 22 + 3, x \geq 22 + 3

各区間の不等式を解く

解なし or x = \frac{15 - 249}{4} or (x = 1 or x = \frac{7}{2}) or x = \frac{15 + 249}{4}

x = \frac{15 - 249}{4} or x = 1 or x = \frac{7}{2} or x = \frac{15 + 249}{4}