sqrt(3)x^2+pix-4<0

Lösung

3 x^{2} + π x - 4 < 0

\frac{- 16 3 + π ^{2} - π}{2 3} < x < \frac{16 3 + π ^{2} - π}{2 3}

Intervall-Notation

(\frac{- 16 3 + π ^{2} - π}{2 3}, \frac{16 3 + π ^{2} - π}{2 3})

Dezimale

- 2.67660 \dots < x < 0.86280 \dots

Schritte zur Lösung

3 x^{2} + π x - 4 < 0

Vervollständige das Quadrat

3 x^{2} + π x - 4 : 3 (x + \frac{π}{2 3})^{2} - 4 - \frac{π ^{2}}{4 3}

3 (x + \frac{π}{2 3})^{2} - 4 - \frac{π ^{2}}{4 3} < 0

Faktor

3 x^{2} + π x - 4 : (x - \frac{- π + π ^{2} + 16 3}{2 3}) (x - \frac{- π - π ^{2} + 16 3}{2 3})

3 (x + \frac{π}{2 3})^{2} - 4 - \frac{π ^{2}}{4 3} < 0

Identifiziere die Intervalle

\frac{- 16 3 + π ^{2} - π}{2 3} < x < \frac{16 3 + π ^{2} - π}{2 3}

\frac{2 x}{5} > 1

x^{2} + 2 x = 0

2 x^{2} - 3 x = 14

s + 1 \leq 3

s im pl i f y (\frac{1}{9})^{\frac{- 1}{2}}