Whakaoti i te t^2-3t-4=0

$$t ^ { 2 } - 3 t - 4 = 0$$

Whakaoti mō t

$$t=-1$$

$$t=4$$

Ngā Upane Whakamahi Tauwehe

Hei whakaoti i te whārite, whakatauwehea te t^{2}-3t-4 mā te whakamahi i te tātai t^{2}+\left(a+b\right)t+ab=\left(t+a\right)\left(t+b\right). Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-3 ab=-4$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. 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.

$$a=-4 b=1$$

Me tuhi anō te kīanga whakatauwehe \left(t+a\right)\left(t+b\right) mā ngā uara i tātaihia.

$$\left(t-4\right)\left(t+1\right)$$

Hei kimi otinga whārite, me whakaoti te t-4=0 me te t+1=0.

$$t=4 t=-1$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

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

$$a+b=-3 ab=1\left(-4\right)=-4$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. 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 t^{2}-3t-4 hei \left(t^{2}-4t\right)+\left(t-4\right).

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

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

$$t\left(t-4\right)+t-4$$

Whakatauwehea atu te kīanga pātahi t-4 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(t-4\right)\left(t+1\right)$$

Hei kimi otinga whārite, me whakaoti te t-4=0 me te t+1=0.

$$t=4 t=-1$$

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.

$$t^{2}-3t-4=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -3 mō b, me -4 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Pūrua -3.

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

Whakareatia -4 ki te -4.

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

Tāpiri 9 ki te 16.

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

Tuhia te pūtakerua o te 25.

$$t=\frac{-\left(-3\right)±5}{2}$$

Ko te tauaro o -3 ko 3.

$$t=\frac{3±5}{2}$$

Nā, me whakaoti te whārite t=\frac{3±5}{2} ina he tāpiri te ±. Tāpiri 3 ki te 5.

$$t=\frac{8}{2}$$

Whakawehe 8 ki te 2.

$$t=4$$

Nā, me whakaoti te whārite t=\frac{3±5}{2} ina he tango te ±. Tango 5 mai i 3.

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

Whakawehe -2 ki te 2.

$$t=-1$$

Kua oti te whārite te whakatau.

$$t=4 t=-1$$

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.

$$t^{2}-3t-4=0$$

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

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

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

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

Tango -4 mai i 0.

$$t^{2}-3t=4$$

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

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

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

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

Tāpiri 4 ki te \frac{9}{4}.

$$t^{2}-3t+\frac{9}{4}=\frac{25}{4}$$

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

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

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

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

Whakarūnātia.

$$t-\frac{3}{2}=\frac{5}{2} t-\frac{3}{2}=-\frac{5}{2}$$

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

$$t=4 t=-1$$