Whakaoti i te -3x^2-2x+1=0

$$- 3 x ^ { 2 } - 2 x + 1 = 0$$

Whakaoti mō x

$$x=\frac{1}{3}\approx 0.333333333$$

$$x=-1$$

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 -3x^{2}+ax+bx+1. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-2 ab=-3=-3$$

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.

$$a=1 b=-3$$

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

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

Tauwehea te -x i te tuatahi me te -1 i te rōpū tuarua.

$$-x\left(3x-1\right)-\left(3x-1\right)$$

Whakatauwehea atu te kīanga pātahi 3x-1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(3x-1\right)\left(-x-1\right)$$

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

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

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.

$$-3x^{2}-2x+1=0$$

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

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

Pūrua -2.

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

Whakareatia -4 ki te -3.

$$x=\frac{-\left(-2\right)±\sqrt{4+12}}{2\left(-3\right)}$$

Tāpiri 4 ki te 12.

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

Tuhia te pūtakerua o te 16.

$$x=\frac{-\left(-2\right)±4}{2\left(-3\right)}$$

Ko te tauaro o -2 ko 2.

$$x=\frac{2±4}{2\left(-3\right)}$$

Whakareatia 2 ki te -3.

$$x=\frac{2±4}{-6}$$

Nā, me whakaoti te whārite x=\frac{2±4}{-6} ina he tāpiri te ±. Tāpiri 2 ki te 4.

$$x=\frac{6}{-6}$$

Whakawehe 6 ki te -6.

$$x=-1$$

Nā, me whakaoti te whārite x=\frac{2±4}{-6} ina he tango te ±. Tango 4 mai i 2.

$$x=-\frac{2}{-6}$$

Whakahekea te hautanga \frac{-2}{-6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=\frac{1}{3}$$

Kua oti te whārite te whakatau.

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

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.

$$-3x^{2}-2x+1=0$$

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

$$-3x^{2}-2x+1-1=-1$$

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

$$-3x^{2}-2x=-1$$

Whakawehea ngā taha e rua ki te -3.

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

Mā te whakawehe ki te -3 ka wetekia te whakareanga ki te -3.

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

Whakawehe -2 ki te -3.

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

Whakawehe -1 ki te -3.

$$x^{2}+\frac{2}{3}x=\frac{1}{3}$$

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.

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

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

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

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.

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

Tauwehea x^{2}+\frac{2}{3}x+\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}.

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

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

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

Whakarūnātia.

$$x+\frac{1}{3}=\frac{2}{3} x+\frac{1}{3}=-\frac{2}{3}$$

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

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