Whakaoti i te -9x^2+12x-3=0

$$- 9 x ^ { 2 } + 12 x - 3 = 0$$

Whakaoti mō x

$$x=1$$

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

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Whakawehea ngā taha e rua ki te 3.

$$-3x^{2}+4x-1=0$$

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.

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

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. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=3 b=1$$

Tuhia anō te -3x^{2}+4x-1 hei \left(-3x^{2}+3x\right)+\left(x-1\right).

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

Tauwehea te 3x i te tuatahi me te -1 i te rōpū tuarua.

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

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

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

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

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

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.

$$-9x^{2}+12x-3=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -9 mō a, 12 mō b, me -3 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\left(-9\right)\left(-3\right)}}{2\left(-9\right)}$$

Pūrua 12.

$$x=\frac{-12±\sqrt{144-4\left(-9\right)\left(-3\right)}}{2\left(-9\right)}$$

Whakareatia -4 ki te -9.

$$x=\frac{-12±\sqrt{144+36\left(-3\right)}}{2\left(-9\right)}$$

Whakareatia 36 ki te -3.

$$x=\frac{-12±\sqrt{144-108}}{2\left(-9\right)}$$

Tāpiri 144 ki te -108.

$$x=\frac{-12±\sqrt{36}}{2\left(-9\right)}$$

Tuhia te pūtakerua o te 36.

$$x=\frac{-12±6}{2\left(-9\right)}$$

Whakareatia 2 ki te -9.

$$x=\frac{-12±6}{-18}$$

Nā, me whakaoti te whārite x=\frac{-12±6}{-18} ina he tāpiri te ±. Tāpiri -12 ki te 6.

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

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

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

Nā, me whakaoti te whārite x=\frac{-12±6}{-18} ina he tango te ±. Tango 6 mai i -12.

$$x=-\frac{18}{-18}$$

Whakawehe -18 ki te -18.

$$x=1$$

Kua oti te whārite te whakatau.

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

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.

$$-9x^{2}+12x-3=0$$

Me tāpiri 3 ki ngā taha e rua o te whārite.

$$-9x^{2}+12x-3-\left(-3\right)=-\left(-3\right)$$

Mā te tango i te -3 i a ia ake anō ka toe ko te 0.

$$-9x^{2}+12x=-\left(-3\right)$$

Tango -3 mai i 0.

$$-9x^{2}+12x=3$$

Whakawehea ngā taha e rua ki te -9.

$$\frac{-9x^{2}+12x}{-9}=\frac{3}{-9}$$

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

$$x^{2}+\frac{12}{-9}x=\frac{3}{-9}$$

Whakahekea te hautanga \frac{12}{-9} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 3.

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

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

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

Whakawehea te -\frac{4}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{2}{3}. Nā, tāpiria te pūrua o te -\frac{2}{3} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

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

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

Tāpiri -\frac{1}{3} ki te \frac{4}{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.

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

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

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

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

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

Whakarūnātia.

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

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

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