solvefor x,sqrt(x^2+x+1)=z-4

Lösung

löse nach

x, x^{2} + x + 1 = z - 4

x = \frac{- 1 + 4 z ^{2} - 32 z + 61}{2} {z \geq \frac{8 + 3}{2}}, x = \frac{- 1 - 4 z ^{2} - 32 z + 61}{2} {z \geq \frac{8 + 3}{2}}

Schritte zur Lösung

x^{2} + x + 1 = z - 4

Quadriere beide Seiten:

x^{2} + x + 1 = z^{2} - 8 z + 16

x^{2} + x + 1 = z^{2} - 8 z + 16

Löse

x^{2} + x + 1 = z^{2} - 8 z + 16 : x = \frac{- 1 + 4 z ^{2} - 32 z + 61}{2}, x = \frac{- 1 - 4 z ^{2} - 32 z + 61}{2}

x = \frac{- 1 + 4 z ^{2} - 32 z + 61}{2}, x = \frac{- 1 - 4 z ^{2} - 32 z + 61}{2}

Überprüfe die Lösungen:

x = \frac{- 1 + 4 z ^{2} - 32 z + 61}{2} {z \geq \frac{8 + 3}{2}}, x = \frac{- 1 - 4 z ^{2} - 32 z + 61}{2} {z \geq \frac{8 + 3}{2}}

Die Lösungen sind

x = \frac{- 1 + 4 z ^{2} - 32 z + 61}{2} {z \geq \frac{8 + 3}{2}}, x = \frac{- 1 - 4 z ^{2} - 32 z + 61}{2} {z \geq \frac{8 + 3}{2}}

so l v e f or y, \frac{s}{t} = v^{2} - y^{2}

x + 4 - x^{2} + 5 x + 4 = 0

3 x^{2} + 2 5 x - 5 = 0

c^{7.2} = 36

so l v e f or x, x^{\frac{1}{2}} + y^{\frac{1}{2}} = a^{\frac{1}{2}}