Whakaoti i te 10t(5t-1)=0

$$10 t ( 5 t - 1 ) = 0$$

Whakaoti mō t

$$t=\frac{1}{5}=0.2$$

$$t=0$$

Ngā Upane Whakamahi Tauwehe

Whakamahia te āhuatanga tohatoha hei whakarea te 10t ki te 5t-1.

$$50t^{2}-10t=0$$

Tauwehea te t.

$$t\left(50t-10\right)=0$$

Hei kimi otinga whārite, me whakaoti te t=0 me te 50t-10=0.

$$t=0 t=\frac{1}{5}$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 10t ki te 5t-1.

$$50t^{2}-10t=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 50 mō a, -10 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}}}{2\times 50}$$

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

$$t=\frac{-\left(-10\right)±10}{2\times 50}$$

Ko te tauaro o -10 ko 10.

$$t=\frac{10±10}{2\times 50}$$

Whakareatia 2 ki te 50.

$$t=\frac{10±10}{100}$$

Nā, me whakaoti te whārite t=\frac{10±10}{100} ina he tāpiri te ±. Tāpiri 10 ki te 10.

$$t=\frac{20}{100}$$

Whakahekea te hautanga \frac{20}{100} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 20.

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

Nā, me whakaoti te whārite t=\frac{10±10}{100} ina he tango te ±. Tango 10 mai i 10.

$$t=\frac{0}{100}$$

Whakawehe 0 ki te 100.

$$t=0$$

Kua oti te whārite te whakatau.

$$t=\frac{1}{5} t=0$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 10t ki te 5t-1.

$$50t^{2}-10t=0$$

Whakawehea ngā taha e rua ki te 50.

$$\frac{50t^{2}-10t}{50}=\frac{0}{50}$$

Mā te whakawehe ki te 50 ka wetekia te whakareanga ki te 50.

$$t^{2}+\left(-\frac{10}{50}\right)t=\frac{0}{50}$$

Whakahekea te hautanga \frac{-10}{50} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 10.

$$t^{2}-\frac{1}{5}t=\frac{0}{50}$$

Whakawehe 0 ki te 50.

$$t^{2}-\frac{1}{5}t=0$$

Whakawehea te -\frac{1}{5}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{10}. Nā, tāpiria te pūrua o te -\frac{1}{10} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$t^{2}-\frac{1}{5}t+\left(-\frac{1}{10}\right)^{2}=\left(-\frac{1}{10}\right)^{2}$$

Pūruatia -\frac{1}{10} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$t^{2}-\frac{1}{5}t+\frac{1}{100}=\frac{1}{100}$$

Tauwehea t^{2}-\frac{1}{5}t+\frac{1}{100}. 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}.

$$\left(t-\frac{1}{10}\right)^{2}=\frac{1}{100}$$

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

$$\sqrt{\left(t-\frac{1}{10}\right)^{2}}=\sqrt{\frac{1}{100}}$$

Whakarūnātia.

$$t-\frac{1}{10}=\frac{1}{10} t-\frac{1}{10}=-\frac{1}{10}$$

Me tāpiri \frac{1}{10} ki ngā taha e rua o te whārite.

$$t=\frac{1}{5} t=0$$