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

Lösung

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}

Dezimale

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

Schritte zur Lösung

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

Finde positive und negative Intervalle

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

Löse die Ungleichung für jedes Intervall

keine L \overset{o}{¨} sung 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}

(4 y + 2)^{2} = 4

s im pl i f y \frac{7 ^{- 6}}{7 ^{- 2}}

f a c t or x^{2} + 16 x + 64

13 = \frac{5}{2} (5 x + 4)

- 16 x^{2} + 10 x + 15 = 0