(x-9)^2>= sqrt(2)(x-9)

Lösung

(x - 9)^{2} \geq 2 (x - 9)

x \leq - \frac{1}{2} - \frac{- 18 - 2}{2} or x \geq \frac{1}{2} - \frac{- 18 - 2}{2}

Intervall-Notation

(- \infty, - \frac{1}{2} - \frac{- 18 - 2}{2}] \cup [\frac{1}{2} - \frac{- 18 - 2}{2}, \infty)

Dezimale

x \leq 9 or x \geq 10.41421 \dots

Schritte zur Lösung

(x - 9)^{2} \geq 2 (x - 9)

Rewrite in standard form

x^{2} - 18 x + 81 + 92 - 2 x \geq 0

Vervollständige das Quadrat

x^{2} - 18 x + 81 + 92 - 2 x : (x + \frac{- 18 - 2}{2})^{2} - \frac{163}{2} + 81

(x + \frac{- 18 - 2}{2})^{2} - \frac{163}{2} + 81 \geq 0

Identifiziere die Intervalle

x \leq - \frac{1}{2} - \frac{- 18 - 2}{2} or x \geq \frac{1}{2} - \frac{- 18 - 2}{2}

- 3 n < 7 n - 20 < 3 n

f a c t or 36 v^{2} - 25

s im pl i f y \frac{\frac{1}{3}}{\frac{1}{3}}

(x + 10) (x - 3) = 0

80 + a^{2} = 18 a