Whakaoti i te -6x^2=3x-3

$$- 6 x ^ { 2 } = 3 x - 3$$

Whakaoti mō x

$$x=\frac{1}{2}=0.5$$

$$x=-1$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tangohia te 3x mai i ngā taha e rua.

$$-6x^{2}-3x=-3$$

Me tāpiri te 3 ki ngā taha e rua.

$$-6x^{2}-3x+3=0$$

Whakawehea ngā taha e rua ki te 3.

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

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

$$a+b=-1 ab=-2=-2$$

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.

$$a=1 b=-2$$

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

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

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

$$-x\left(2x-1\right)-\left(2x-1\right)$$

Whakatauwehea atu te kīanga pātahi 2x-1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2x-1\right)\left(-x-1\right)$$

Hei kimi otinga whārite, me whakaoti te 2x-1=0 me te -x-1=0.

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 3x mai i ngā taha e rua.

$$-6x^{2}-3x=-3$$

Me tāpiri te 3 ki ngā taha e rua.

$$-6x^{2}-3x+3=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -6 mō a, -3 mō b, me 3 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-6\right)\times 3}}{2\left(-6\right)}$$

Pūrua -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-6\right)\times 3}}{2\left(-6\right)}$$

Whakareatia -4 ki te -6.

$$x=\frac{-\left(-3\right)±\sqrt{9+24\times 3}}{2\left(-6\right)}$$

Whakareatia 24 ki te 3.

$$x=\frac{-\left(-3\right)±\sqrt{9+72}}{2\left(-6\right)}$$

Tāpiri 9 ki te 72.

$$x=\frac{-\left(-3\right)±\sqrt{81}}{2\left(-6\right)}$$

Tuhia te pūtakerua o te 81.

$$x=\frac{-\left(-3\right)±9}{2\left(-6\right)}$$

Ko te tauaro o -3 ko 3.

$$x=\frac{3±9}{2\left(-6\right)}$$

Whakareatia 2 ki te -6.

$$x=\frac{3±9}{-12}$$

Nā, me whakaoti te whārite x=\frac{3±9}{-12} ina he tāpiri te ±. Tāpiri 3 ki te 9.

$$x=\frac{12}{-12}$$

Whakawehe 12 ki te -12.

$$x=-1$$

Nā, me whakaoti te whārite x=\frac{3±9}{-12} ina he tango te ±. Tango 9 mai i 3.

$$x=-\frac{6}{-12}$$

Whakahekea te hautanga \frac{-6}{-12} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

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

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 3x mai i ngā taha e rua.

$$-6x^{2}-3x=-3$$

Whakawehea ngā taha e rua ki te -6.

$$\frac{-6x^{2}-3x}{-6}=-\frac{3}{-6}$$

Mā te whakawehe ki te -6 ka wetekia te whakareanga ki te -6.

$$x^{2}+\left(-\frac{3}{-6}\right)x=-\frac{3}{-6}$$

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

$$x^{2}+\frac{1}{2}x=-\frac{3}{-6}$$

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

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

Whakawehea te \frac{1}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{4}. Nā, tāpiria te pūrua o te \frac{1}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia \frac{1}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{1}{2}+\frac{1}{16}$$

Tāpiri \frac{1}{2} ki te \frac{1}{16} 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.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{9}{16}$$

Tauwehea x^{2}+\frac{1}{2}x+\frac{1}{16}. 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}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{9}{16}$$

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

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Whakarūnātia.

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

Me tango \frac{1}{4} mai i ngā taha e rua o te whārite.

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