Whakaoti i te 9t^2+6t+1=0

$$9 t ^ { 2 } + 6 t + 1 = 0$$

Whakaoti mō t

$$t=-\frac{1}{3}\approx -0.333333333$$

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 9t^{2}+at+bt+1. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=6 ab=9\times 1=9$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōrunga te a+b, he tōrunga hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 9.

$$1,9 3,3$$

Tātaihia te tapeke mō ia takirua.

$$1+9=10 3+3=6$$

Ko te otinga te takirua ka hoatu i te tapeke 6.

$$a=3 b=3$$

Tuhia anō te 9t^{2}+6t+1 hei \left(9t^{2}+3t\right)+\left(3t+1\right).

$$\left(9t^{2}+3t\right)+\left(3t+1\right)$$

Whakatauwehea atu 3t i te 9t^{2}+3t.

$$3t\left(3t+1\right)+3t+1$$

Whakatauwehea atu te kīanga pātahi 3t+1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(3t+1\right)\left(3t+1\right)$$

Tuhia anōtia hei pūrua huarua.

$$\left(3t+1\right)^{2}$$

Hei kimi i te otinga whārite, whakaotia te 3t+1=0.

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

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.

$$9t^{2}+6t+1=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 9 mō a, 6 mō b, me 1 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-6±\sqrt{6^{2}-4\times 9}}{2\times 9}$$

Pūrua 6.

$$t=\frac{-6±\sqrt{36-4\times 9}}{2\times 9}$$

Whakareatia -4 ki te 9.

$$t=\frac{-6±\sqrt{36-36}}{2\times 9}$$

Tāpiri 36 ki te -36.

$$t=\frac{-6±\sqrt{0}}{2\times 9}$$

Tuhia te pūtakerua o te 0.

$$t=-\frac{6}{2\times 9}$$

Whakareatia 2 ki te 9.

$$t=-\frac{6}{18}$$

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

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

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.

$$9t^{2}+6t+1=0$$

Me tango 1 mai i ngā taha e rua o te whārite.

$$9t^{2}+6t+1-1=-1$$

Mā te tango i te 1 i a ia ake anō ka toe ko te 0.

$$9t^{2}+6t=-1$$

Whakawehea ngā taha e rua ki te 9.

$$\frac{9t^{2}+6t}{9}=-\frac{1}{9}$$

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

$$t^{2}+\frac{6}{9}t=-\frac{1}{9}$$

Whakahekea te hautanga \frac{6}{9} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 3.

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

Whakawehea te \frac{2}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{3}. Nā, tāpiria te pūrua o te \frac{1}{3} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$t^{2}+\frac{2}{3}t+\left(\frac{1}{3}\right)^{2}=-\frac{1}{9}+\left(\frac{1}{3}\right)^{2}$$

Pūruatia \frac{1}{3} mā te pūrua i te taurunga me te tauraro o te hautanga.

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

Tāpiri -\frac{1}{9} ki te \frac{1}{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.

$$t^{2}+\frac{2}{3}t+\frac{1}{9}=0$$

Tauwehea t^{2}+\frac{2}{3}t+\frac{1}{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}.

$$\left(t+\frac{1}{3}\right)^{2}=0$$

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

$$\sqrt{\left(t+\frac{1}{3}\right)^{2}}=\sqrt{0}$$

Whakarūnātia.

$$t+\frac{1}{3}=0 t+\frac{1}{3}=0$$

Me tango \frac{1}{3} mai i ngā taha e rua o te whārite.

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

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

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