Whakaoti i te 9t^2-12t=-4

$$9 t ^ { 2 } - 12 t = - 4$$

Whakaoti mō t

$$t=\frac{2}{3}\approx 0.666666667$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Me tāpiri te 4 ki ngā taha e rua.

$$9t^{2}-12t+4=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 9t^{2}+at+bt+4. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-12 ab=9\times 4=36$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōraro te a+b, he tōraro hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 36.

$$-1,-36 -2,-18 -3,-12 -4,-9 -6,-6$$

Tātaihia te tapeke mō ia takirua.

$$-1-36=-37 -2-18=-20 -3-12=-15 -4-9=-13 -6-6=-12$$

Ko te otinga te takirua ka hoatu i te tapeke -12.

$$a=-6 b=-6$$

Tuhia anō te 9t^{2}-12t+4 hei \left(9t^{2}-6t\right)+\left(-6t+4\right).

$$\left(9t^{2}-6t\right)+\left(-6t+4\right)$$

Tauwehea te 3t i te tuatahi me te -2 i te rōpū tuarua.

$$3t\left(3t-2\right)-2\left(3t-2\right)$$

Whakatauwehea atu te kīanga pātahi 3t-2 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(3t-2\right)\left(3t-2\right)$$

Tuhia anōtia hei pūrua huarua.

$$\left(3t-2\right)^{2}$$

Hei kimi i te otinga whārite, whakaotia te 3t-2=0.

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

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.

$$9t^{2}-12t=-4$$

Me tāpiri 4 ki ngā taha e rua o te whārite.

$$9t^{2}-12t-\left(-4\right)=-4-\left(-4\right)$$

Mā te tango i te -4 i a ia ake anō ka toe ko te 0.

$$9t^{2}-12t-\left(-4\right)=0$$

Tango -4 mai i 0.

$$9t^{2}-12t+4=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 9 mō a, -12 mō b, me 4 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 9\times 4}}{2\times 9}$$

Pūrua -12.

$$t=\frac{-\left(-12\right)±\sqrt{144-4\times 9\times 4}}{2\times 9}$$

Whakareatia -4 ki te 9.

$$t=\frac{-\left(-12\right)±\sqrt{144-36\times 4}}{2\times 9}$$

Whakareatia -36 ki te 4.

$$t=\frac{-\left(-12\right)±\sqrt{144-144}}{2\times 9}$$

Tāpiri 144 ki te -144.

$$t=\frac{-\left(-12\right)±\sqrt{0}}{2\times 9}$$

Tuhia te pūtakerua o te 0.

$$t=-\frac{-12}{2\times 9}$$

Ko te tauaro o -12 ko 12.

$$t=\frac{12}{2\times 9}$$

Whakareatia 2 ki te 9.

$$t=\frac{12}{18}$$

Whakahekea te hautanga \frac{12}{18} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

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

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.

$$9t^{2}-12t=-4$$

Whakawehea ngā taha e rua ki te 9.

$$\frac{9t^{2}-12t}{9}=-\frac{4}{9}$$

Mā te whakawehe ki te 9 ka wetekia te whakareanga ki te 9.

$$t^{2}+\left(-\frac{12}{9}\right)t=-\frac{4}{9}$$

Whakahekea te hautanga \frac{-12}{9} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 3.

$$t^{2}-\frac{4}{3}t=-\frac{4}{9}$$

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

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

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

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

Tāpiri -\frac{4}{9} ki te \frac{4}{9} mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$t^{2}-\frac{4}{3}t+\frac{4}{9}=0$$

Tauwehea t^{2}-\frac{4}{3}t+\frac{4}{9}. 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}.

$$\left(t-\frac{2}{3}\right)^{2}=0$$

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

$$\sqrt{\left(t-\frac{2}{3}\right)^{2}}=\sqrt{0}$$

Whakarūnātia.

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

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

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

Kua oti te whārite te whakatau. He ōrite ngā whakatau.

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