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

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

Whakaoti mō x

$$x=-1$$

$$x=\frac{2}{3}\approx 0.666666667$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tangohia te 2x^{2} mai i ngā taha e rua.

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

Pahekotia te -x^{2} me -2x^{2}, ka -3x^{2}.

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

Tangohia te -1 mai i ngā taha e rua.

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

Ko te tauaro o -1 ko 1.

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

Pahekotia te 1 me 1, ka 2\times 1.

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

Tangohia te x mai i ngā taha e rua.

$$2\times 1-3x^{2}-x=0$$

Whakareatia te 2 ki te 1, ka 2.

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

Hurinahatia te pūrau ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

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

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

$$a+b=-1 ab=-3\times 2=-6$$

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. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -6.

$$1,-6 2,-3$$

Tātaihia te tapeke mō ia takirua.

$$1-6=-5 2-3=-1$$

Ko te otinga te takirua ka hoatu i te tapeke -1.

$$a=2 b=-3$$

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

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

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

$$-x\left(3x-2\right)-\left(3x-2\right)$$

Whakatauwehea atu te kīanga pātahi 3x-2 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(3x-2\right)\left(-x-1\right)$$

Hei kimi otinga whārite, me whakaoti te 3x-2=0 me te -x-1=0.

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 2x^{2} mai i ngā taha e rua.

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

Pahekotia te -x^{2} me -2x^{2}, ka -3x^{2}.

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

Tangohia te -1 mai i ngā taha e rua.

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

Ko te tauaro o -1 ko 1.

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

Pahekotia te 1 me 1, ka 2\times 1.

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

Tangohia te x mai i ngā taha e rua.

$$2\times 1-3x^{2}-x=0$$

Whakareatia te 2 ki te 1, ka 2.

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

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.

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

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

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-3\right)\times 2}}{2\left(-3\right)}$$

Whakareatia -4 ki te -3.

$$x=\frac{-\left(-1\right)±\sqrt{1+12\times 2}}{2\left(-3\right)}$$

Whakareatia 12 ki te 2.

$$x=\frac{-\left(-1\right)±\sqrt{1+24}}{2\left(-3\right)}$$

Tāpiri 1 ki te 24.

$$x=\frac{-\left(-1\right)±\sqrt{25}}{2\left(-3\right)}$$

Tuhia te pūtakerua o te 25.

$$x=\frac{-\left(-1\right)±5}{2\left(-3\right)}$$

Ko te tauaro o -1 ko 1.

$$x=\frac{1±5}{2\left(-3\right)}$$

Whakareatia 2 ki te -3.

$$x=\frac{1±5}{-6}$$

Nā, me whakaoti te whārite x=\frac{1±5}{-6} ina he tāpiri te ±. Tāpiri 1 ki te 5.

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

Whakawehe 6 ki te -6.

$$x=-1$$

Nā, me whakaoti te whārite x=\frac{1±5}{-6} ina he tango te ±. Tango 5 mai i 1.

$$x=-\frac{4}{-6}$$

Whakahekea te hautanga \frac{-4}{-6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 2x^{2} mai i ngā taha e rua.

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

Pahekotia te -x^{2} me -2x^{2}, ka -3x^{2}.

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

Tangohia te x mai i ngā taha e rua.

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

Tangohia te 1 mai i ngā taha e rua.

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

Tangohia te 1 i te -1, ka -2.

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

Whakawehea ngā taha e rua ki te -3.

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

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

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

Whakawehe -1 ki te -3.

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

Whakawehe -2 ki te -3.

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

Whakawehea te \frac{1}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{6}. Nā, tāpiria te pūrua o te \frac{1}{6} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{1}{3}x+\left(\frac{1}{6}\right)^{2}=\frac{2}{3}+\left(\frac{1}{6}\right)^{2}$$

Pūruatia \frac{1}{6} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{2}{3}+\frac{1}{36}$$

Tāpiri \frac{2}{3} ki te \frac{1}{36} 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.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{25}{36}$$

Tauwehea x^{2}+\frac{1}{3}x+\frac{1}{36}. 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}.

$$\left(x+\frac{1}{6}\right)^{2}=\frac{25}{36}$$

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

$$\sqrt{\left(x+\frac{1}{6}\right)^{2}}=\sqrt{\frac{25}{36}}$$

Whakarūnātia.

$$x+\frac{1}{6}=\frac{5}{6} x+\frac{1}{6}=-\frac{5}{6}$$

Me tango \frac{1}{6} mai i ngā taha e rua o te whārite.

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