Whakaoti i te 4t^2-10t=0

$$4 t ^ { 2 } - 10 t = 0$$

Whakaoti mō t

$$t = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

$$t=0$$

Ngā Upane Whakamahi Tauwehe

Tauwehea te t.

$$t\left(4t-10\right)=0$$

Hei kimi otinga whārite, me whakaoti te t=0 me te 4t-10=0.

$$t=0 t=\frac{5}{2}$$

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.

$$4t^{2}-10t=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 4 mō a, -10 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}}}{2\times 4}$$

Tuhia te pūtakerua o te \left(-10\right)^{2}.

$$t=\frac{-\left(-10\right)±10}{2\times 4}$$

Ko te tauaro o -10 ko 10.

$$t=\frac{10±10}{2\times 4}$$

Whakareatia 2 ki te 4.

$$t=\frac{10±10}{8}$$

Nā, me whakaoti te whārite t=\frac{10±10}{8} ina he tāpiri te ±. Tāpiri 10 ki te 10.

$$t=\frac{20}{8}$$

Whakahekea te hautanga \frac{20}{8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$t=\frac{5}{2}$$

Nā, me whakaoti te whārite t=\frac{10±10}{8} ina he tango te ±. Tango 10 mai i 10.

$$t=\frac{0}{8}$$

Whakawehe 0 ki te 8.

$$t=0$$

Kua oti te whārite te whakatau.

$$t=\frac{5}{2} t=0$$

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.

$$4t^{2}-10t=0$$

Whakawehea ngā taha e rua ki te 4.

$$\frac{4t^{2}-10t}{4}=\frac{0}{4}$$

Mā te whakawehe ki te 4 ka wetekia te whakareanga ki te 4.

$$t^{2}+\left(-\frac{10}{4}\right)t=\frac{0}{4}$$

Whakahekea te hautanga \frac{-10}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$t^{2}-\frac{5}{2}t=\frac{0}{4}$$

Whakawehe 0 ki te 4.

$$t^{2}-\frac{5}{2}t=0$$

Whakawehea te -\frac{5}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{5}{4}. Nā, tāpiria te pūrua o te -\frac{5}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$t^{2}-\frac{5}{2}t+\left(-\frac{5}{4}\right)^{2}=\left(-\frac{5}{4}\right)^{2}$$

Pūruatia -\frac{5}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$t^{2}-\frac{5}{2}t+\frac{25}{16}=\frac{25}{16}$$

Tauwehea t^{2}-\frac{5}{2}t+\frac{25}{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}.

$$\left(t-\frac{5}{4}\right)^{2}=\frac{25}{16}$$

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

$$\sqrt{\left(t-\frac{5}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Whakarūnātia.

$$t-\frac{5}{4}=\frac{5}{4} t-\frac{5}{4}=-\frac{5}{4}$$

Me tāpiri \frac{5}{4} ki ngā taha e rua o te whārite.

$$t=\frac{5}{2} t=0$$