Whakaoti i te 8q^2-8q=0

$$8 q ^ { 2 } - 8 q = 0$$

Whakaoti mō q

$$q=1$$

$$q=0$$

Ngā Upane Whakamahi Tauwehe

Tauwehea te q.

$$q\left(8q-8\right)=0$$

Hei kimi otinga whārite, me whakaoti te q=0 me te 8q-8=0.

$$q=0 q=1$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.

$$8q^{2}-8q=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 8 mō a, -8 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$q=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}}}{2\times 8}$$

Tuhia te pūtakerua o te \left(-8\right)^{2}.

$$q=\frac{-\left(-8\right)±8}{2\times 8}$$

Ko te tauaro o -8 ko 8.

$$q=\frac{8±8}{2\times 8}$$

Whakareatia 2 ki te 8.

$$q=\frac{8±8}{16}$$

Nā, me whakaoti te whārite q=\frac{8±8}{16} ina he tāpiri te ±. Tāpiri 8 ki te 8.

$$q=\frac{16}{16}$$

Whakawehe 16 ki te 16.

$$q=1$$

Nā, me whakaoti te whārite q=\frac{8±8}{16} ina he tango te ±. Tango 8 mai i 8.

$$q=\frac{0}{16}$$

Whakawehe 0 ki te 16.

$$q=0$$

Kua oti te whārite te whakatau.

$$q=1 q=0$$

Ngā Upane mō te Whakaoti i te Pūrua

Ko ngā whārite pūrua pēnei i tēnei nā ka taea te whakaoti mā te whakaoti i te pūrua. Hei whakaoti i te pūrua, ko te whārite me mātua tuhi ki te āhua x^{2}+bx=c.

$$8q^{2}-8q=0$$

Whakawehea ngā taha e rua ki te 8.

$$\frac{8q^{2}-8q}{8}=\frac{0}{8}$$

Mā te whakawehe ki te 8 ka wetekia te whakareanga ki te 8.

$$q^{2}+\left(-\frac{8}{8}\right)q=\frac{0}{8}$$

Whakawehe -8 ki te 8.

$$q^{2}-q=\frac{0}{8}$$

Whakawehe 0 ki te 8.

$$q^{2}-q=0$$

Whakawehea te -1, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{2}. Nā, tāpiria te pūrua o te -\frac{1}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$q^{2}-q+\left(-\frac{1}{2}\right)^{2}=\left(-\frac{1}{2}\right)^{2}$$

Pūruatia -\frac{1}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$q^{2}-q+\frac{1}{4}=\frac{1}{4}$$

Tauwehea q^{2}-q+\frac{1}{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}.

$$\left(q-\frac{1}{2}\right)^{2}=\frac{1}{4}$$

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

$$\sqrt{\left(q-\frac{1}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Whakarūnātia.

$$q-\frac{1}{2}=\frac{1}{2} q-\frac{1}{2}=-\frac{1}{2}$$

Me tāpiri \frac{1}{2} ki ngā taha e rua o te whārite.

$$q=1 q=0$$