Whakaoti i te 2+3t=2t^2

$$2 + 3 t = 2 t ^ { 2 }$$

Whakaoti mō t

$$t=2$$

$$t=-\frac{1}{2}=-0.5$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tangohia te 2t^{2} mai i ngā taha e rua.

$$2+3t-2t^{2}=0$$

Hurinahatia te pūrau ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

$$-2t^{2}+3t+2=0$$

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

$$a+b=3 ab=-2\times 2=-4$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōrunga te a+b, he nui ake te uara pū o te tau tōrunga i tō te tōraro. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -4.

$$-1,4 -2,2$$

Tātaihia te tapeke mō ia takirua.

$$-1+4=3 -2+2=0$$

Ko te otinga te takirua ka hoatu i te tapeke 3.

$$a=4 b=-1$$

Tuhia anō te -2t^{2}+3t+2 hei \left(-2t^{2}+4t\right)+\left(-t+2\right).

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

Whakatauwehea atu 2t i te -2t^{2}+4t.

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

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

$$\left(-t+2\right)\left(2t+1\right)$$

Hei kimi otinga whārite, me whakaoti te -t+2=0 me te 2t+1=0.

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 2t^{2} mai i ngā taha e rua.

$$2+3t-2t^{2}=0$$

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.

$$-2t^{2}+3t+2=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -2 mō a, 3 mō b, me 2 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Pūrua 3.

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

Whakareatia -4 ki te -2.

$$t=\frac{-3±\sqrt{9+8\times 2}}{2\left(-2\right)}$$

Whakareatia 8 ki te 2.

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

Tāpiri 9 ki te 16.

$$t=\frac{-3±\sqrt{25}}{2\left(-2\right)}$$

Tuhia te pūtakerua o te 25.

$$t=\frac{-3±5}{2\left(-2\right)}$$

Whakareatia 2 ki te -2.

$$t=\frac{-3±5}{-4}$$

Nā, me whakaoti te whārite t=\frac{-3±5}{-4} ina he tāpiri te ±. Tāpiri -3 ki te 5.

$$t=\frac{2}{-4}$$

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

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

Nā, me whakaoti te whārite t=\frac{-3±5}{-4} ina he tango te ±. Tango 5 mai i -3.

$$t=-\frac{8}{-4}$$

Whakawehe -8 ki te -4.

$$t=2$$

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 2t^{2} mai i ngā taha e rua.

$$2+3t-2t^{2}=0$$

Tangohia te 2 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$3t-2t^{2}=-2$$

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.

$$-2t^{2}+3t=-2$$

Whakawehea ngā taha e rua ki te -2.

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

Mā te whakawehe ki te -2 ka wetekia te whakareanga ki te -2.

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

Whakawehe 3 ki te -2.

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

Whakawehe -2 ki te -2.

$$t^{2}-\frac{3}{2}t=1$$

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.

$$t^{2}-\frac{3}{2}t+\left(-\frac{3}{4}\right)^{2}=1+\left(-\frac{3}{4}\right)^{2}$$

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

$$t^{2}-\frac{3}{2}t+\frac{9}{16}=1+\frac{9}{16}$$

Tāpiri 1 ki te \frac{9}{16}.

$$t^{2}-\frac{3}{2}t+\frac{9}{16}=\frac{25}{16}$$

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}.

$$\left(t-\frac{3}{4}\right)^{2}=\frac{25}{16}$$

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

$$\sqrt{\left(t-\frac{3}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Whakarūnātia.

$$t-\frac{3}{4}=\frac{5}{4} t-\frac{3}{4}=-\frac{5}{4}$$

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

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