Whakaoti i te 8t^2-12t+9=9

$$8 { t }^{ 2 } -12t+9 = 9$$

Whakaoti mō t

$$t = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

$$t=0$$

Ngā Upane Whakamahi Tauwehe

Tangohia te 9 mai i ngā taha e rua.

$$8t^{2}-12t+9-9=0$$

Tangohia te 9 i te 9, ka 0.

$$8t^{2}-12t=0$$

Tauwehea te t.

$$t\left(8t-12\right)=0$$

Hei kimi otinga whārite, me whakaoti te t=0 me te 8t-12=0.

$$t=0 t=\frac{3}{2}$$

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.

$$8t^{2}-12t+9=9$$

Me tango 9 mai i ngā taha e rua o te whārite.

$$8t^{2}-12t+9-9=9-9$$

Mā te tango i te 9 i a ia ake anō ka toe ko te 0.

$$8t^{2}-12t+9-9=0$$

Tango 9 mai i 9.

$$8t^{2}-12t=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 8 mō a, -12 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}}}{2\times 8}$$

Tuhia te pūtakerua o te \left(-12\right)^{2}.

$$t=\frac{-\left(-12\right)±12}{2\times 8}$$

Ko te tauaro o -12 ko 12.

$$t=\frac{12±12}{2\times 8}$$

Whakareatia 2 ki te 8.

$$t=\frac{12±12}{16}$$

Nā, me whakaoti te whārite t=\frac{12±12}{16} ina he tāpiri te ±. Tāpiri 12 ki te 12.

$$t=\frac{24}{16}$$

Whakahekea te hautanga \frac{24}{16} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 8.

$$t=\frac{3}{2}$$

Nā, me whakaoti te whārite t=\frac{12±12}{16} ina he tango te ±. Tango 12 mai i 12.

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

Whakawehe 0 ki te 16.

$$t=0$$

Kua oti te whārite te whakatau.

$$t=\frac{3}{2} t=0$$

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.

$$8t^{2}-12t+9=9$$

Me tango 9 mai i ngā taha e rua o te whārite.

$$8t^{2}-12t+9-9=9-9$$

Mā te tango i te 9 i a ia ake anō ka toe ko te 0.

$$8t^{2}-12t=9-9$$

Tango 9 mai i 9.

$$8t^{2}-12t=0$$

Whakawehea ngā taha e rua ki te 8.

$$\frac{8t^{2}-12t}{8}=\frac{0}{8}$$

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

$$t^{2}+\left(-\frac{12}{8}\right)t=\frac{0}{8}$$

Whakahekea te hautanga \frac{-12}{8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$t^{2}-\frac{3}{2}t=\frac{0}{8}$$

Whakawehe 0 ki te 8.

$$t^{2}-\frac{3}{2}t=0$$

Whakawehea te -\frac{3}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{3}{4}. Nā, tāpiria te pūrua o te -\frac{3}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$t^{2}-\frac{3}{2}t+\left(-\frac{3}{4}\right)^{2}=\left(-\frac{3}{4}\right)^{2}$$

Pūruatia -\frac{3}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$t^{2}-\frac{3}{2}t+\frac{9}{16}=\frac{9}{16}$$

Tauwehea t^{2}-\frac{3}{2}t+\frac{9}{16}. 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}.

$$\left(t-\frac{3}{4}\right)^{2}=\frac{9}{16}$$

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

$$\sqrt{\left(t-\frac{3}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Whakarūnātia.

$$t-\frac{3}{4}=\frac{3}{4} t-\frac{3}{4}=-\frac{3}{4}$$

Me tāpiri \frac{3}{4} ki ngā taha e rua o te whārite.

$$t=\frac{3}{2} t=0$$