Whakaoti i te -5z^2-3z-11=-6z^2

$$- 5 z ^ { 2 } - 3 z - 11 = - 6 z ^ { 2 }$$

Whakaoti mō z

$$z = \frac{\sqrt{53} + 3}{2} \approx 5.140054945$$

$$z=\frac{3-\sqrt{53}}{2}\approx -2.140054945$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Me tāpiri te 6z^{2} ki ngā taha e rua.

$$-5z^{2}-3z-11+6z^{2}=0$$

Pahekotia te -5z^{2} me 6z^{2}, ka z^{2}.

$$z^{2}-3z-11=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -3 mō b, me -11 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$z=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-11\right)}}{2}$$

Pūrua -3.

$$z=\frac{-\left(-3\right)±\sqrt{9-4\left(-11\right)}}{2}$$

Whakareatia -4 ki te -11.

$$z=\frac{-\left(-3\right)±\sqrt{9+44}}{2}$$

Tāpiri 9 ki te 44.

$$z=\frac{-\left(-3\right)±\sqrt{53}}{2}$$

Ko te tauaro o -3 ko 3.

$$z=\frac{3±\sqrt{53}}{2}$$

Nā, me whakaoti te whārite z=\frac{3±\sqrt{53}}{2} ina he tāpiri te ±. Tāpiri 3 ki te \sqrt{53}.

$$z=\frac{\sqrt{53}+3}{2}$$

Nā, me whakaoti te whārite z=\frac{3±\sqrt{53}}{2} ina he tango te ±. Tango \sqrt{53} mai i 3.

$$z=\frac{3-\sqrt{53}}{2}$$

Kua oti te whārite te whakatau.

$$z=\frac{\sqrt{53}+3}{2} z=\frac{3-\sqrt{53}}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

Me tāpiri te 6z^{2} ki ngā taha e rua.

$$-5z^{2}-3z-11+6z^{2}=0$$

Pahekotia te -5z^{2} me 6z^{2}, ka z^{2}.

$$z^{2}-3z-11=0$$

Me tāpiri te 11 ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

$$z^{2}-3z=11$$

Whakawehea te -3, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{3}{2}. Nā, tāpiria te pūrua o te -\frac{3}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$z^{2}-3z+\left(-\frac{3}{2}\right)^{2}=11+\left(-\frac{3}{2}\right)^{2}$$

Pūruatia -\frac{3}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$z^{2}-3z+\frac{9}{4}=11+\frac{9}{4}$$

Tāpiri 11 ki te \frac{9}{4}.

$$z^{2}-3z+\frac{9}{4}=\frac{53}{4}$$

Tauwehea z^{2}-3z+\frac{9}{4}. Ko te tikanga pūnoa, ina ko x^{2}+bx+c he pūrua tika pūrua tika pū, ka taea taua mea te tauwehea i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(z-\frac{3}{2}\right)^{2}=\frac{53}{4}$$

Tuhia te pūtakerua o ngā taha e rua o te whārite.

$$\sqrt{\left(z-\frac{3}{2}\right)^{2}}=\sqrt{\frac{53}{4}}$$

Whakarūnātia.

$$z-\frac{3}{2}=\frac{\sqrt{53}}{2} z-\frac{3}{2}=-\frac{\sqrt{53}}{2}$$

Me tāpiri \frac{3}{2} ki ngā taha e rua o te whārite.

$$z=\frac{\sqrt{53}+3}{2} z=\frac{3-\sqrt{53}}{2}$$