Whakaoti i te 21c^2-16c-6=1-2c

$$21 c ^ { 2 } - 16 c - 6 = 1 - 2 c$$

Whakaoti mō c

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

$$c=1$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tangohia te 1 mai i ngā taha e rua.

$$21c^{2}-16c-6-1=-2c$$

Tangohia te 1 i te -6, ka -7.

$$21c^{2}-16c-7=-2c$$

Me tāpiri te 2c ki ngā taha e rua.

$$21c^{2}-16c-7+2c=0$$

Pahekotia te -16c me 2c, ka -14c.

$$21c^{2}-14c-7=0$$

Whakawehea ngā taha e rua ki te 7.

$$3c^{2}-2c-1=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei 3c^{2}+ac+bc-1. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-2 ab=3\left(-1\right)=-3$$

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. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=-3 b=1$$

Tuhia anō te 3c^{2}-2c-1 hei \left(3c^{2}-3c\right)+\left(c-1\right).

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

Whakatauwehea atu 3c i te 3c^{2}-3c.

$$3c\left(c-1\right)+c-1$$

Whakatauwehea atu te kīanga pātahi c-1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(c-1\right)\left(3c+1\right)$$

Hei kimi otinga whārite, me whakaoti te c-1=0 me te 3c+1=0.

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 1 mai i ngā taha e rua.

$$21c^{2}-16c-6-1=-2c$$

Tangohia te 1 i te -6, ka -7.

$$21c^{2}-16c-7=-2c$$

Me tāpiri te 2c ki ngā taha e rua.

$$21c^{2}-16c-7+2c=0$$

Pahekotia te -16c me 2c, ka -14c.

$$21c^{2}-14c-7=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 21 mō a, -14 mō b, me -7 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$c=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 21\left(-7\right)}}{2\times 21}$$

Pūrua -14.

$$c=\frac{-\left(-14\right)±\sqrt{196-4\times 21\left(-7\right)}}{2\times 21}$$

Whakareatia -4 ki te 21.

$$c=\frac{-\left(-14\right)±\sqrt{196-84\left(-7\right)}}{2\times 21}$$

Whakareatia -84 ki te -7.

$$c=\frac{-\left(-14\right)±\sqrt{196+588}}{2\times 21}$$

Tāpiri 196 ki te 588.

$$c=\frac{-\left(-14\right)±\sqrt{784}}{2\times 21}$$

Tuhia te pūtakerua o te 784.

$$c=\frac{-\left(-14\right)±28}{2\times 21}$$

Ko te tauaro o -14 ko 14.

$$c=\frac{14±28}{2\times 21}$$

Whakareatia 2 ki te 21.

$$c=\frac{14±28}{42}$$

Nā, me whakaoti te whārite c=\frac{14±28}{42} ina he tāpiri te ±. Tāpiri 14 ki te 28.

$$c=\frac{42}{42}$$

Whakawehe 42 ki te 42.

$$c=1$$

Nā, me whakaoti te whārite c=\frac{14±28}{42} ina he tango te ±. Tango 28 mai i 14.

$$c=-\frac{14}{42}$$

Whakahekea te hautanga \frac{-14}{42} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 14.

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

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Me tāpiri te 2c ki ngā taha e rua.

$$21c^{2}-16c-6+2c=1$$

Pahekotia te -16c me 2c, ka -14c.

$$21c^{2}-14c-6=1$$

Me tāpiri te 6 ki ngā taha e rua.

$$21c^{2}-14c=1+6$$

Tāpirihia te 1 ki te 6, ka 7.

$$21c^{2}-14c=7$$

Whakawehea ngā taha e rua ki te 21.

$$\frac{21c^{2}-14c}{21}=\frac{7}{21}$$

Mā te whakawehe ki te 21 ka wetekia te whakareanga ki te 21.

$$c^{2}+\left(-\frac{14}{21}\right)c=\frac{7}{21}$$

Whakahekea te hautanga \frac{-14}{21} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 7.

$$c^{2}-\frac{2}{3}c=\frac{7}{21}$$

Whakahekea te hautanga \frac{7}{21} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 7.

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

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.

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

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

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

Tāpiri \frac{1}{3} 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.

$$c^{2}-\frac{2}{3}c+\frac{1}{9}=\frac{4}{9}$$

Tauwehea c^{2}-\frac{2}{3}c+\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}.

$$\left(c-\frac{1}{3}\right)^{2}=\frac{4}{9}$$

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

$$\sqrt{\left(c-\frac{1}{3}\right)^{2}}=\sqrt{\frac{4}{9}}$$

Whakarūnātia.

$$c-\frac{1}{3}=\frac{2}{3} c-\frac{1}{3}=-\frac{2}{3}$$

Me tāpiri \frac{1}{3} ki ngā taha e rua o te whārite.

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