Whakaoti i te 2+y(1-3y)=y(y-3)

$$2 + y ( 1 - 3 y ) = y ( y - 3 )$$

Whakaoti mō y

$$y = \frac{\sqrt{3} + 1}{2} \approx 1.366025404$$

$$y=\frac{1-\sqrt{3}}{2}\approx -0.366025404$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te y ki te 1-3y.

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

Whakamahia te āhuatanga tohatoha hei whakarea te y ki te y-3.

$$2+y-3y^{2}=y^{2}-3y$$

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

$$2+y-3y^{2}-y^{2}=-3y$$

Pahekotia te -3y^{2} me -y^{2}, ka -4y^{2}.

$$2+y-4y^{2}=-3y$$

Me tāpiri te 3y ki ngā taha e rua.

$$2+y-4y^{2}+3y=0$$

Pahekotia te y me 3y, ka 4y.

$$2+4y-4y^{2}=0$$

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.

$$-4y^{2}+4y+2=0$$

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

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

Pūrua 4.

$$y=\frac{-4±\sqrt{16-4\left(-4\right)\times 2}}{2\left(-4\right)}$$

Whakareatia -4 ki te -4.

$$y=\frac{-4±\sqrt{16+16\times 2}}{2\left(-4\right)}$$

Whakareatia 16 ki te 2.

$$y=\frac{-4±\sqrt{16+32}}{2\left(-4\right)}$$

Tāpiri 16 ki te 32.

$$y=\frac{-4±\sqrt{48}}{2\left(-4\right)}$$

Tuhia te pūtakerua o te 48.

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

Whakareatia 2 ki te -4.

$$y=\frac{-4±4\sqrt{3}}{-8}$$

Nā, me whakaoti te whārite y=\frac{-4±4\sqrt{3}}{-8} ina he tāpiri te ±. Tāpiri -4 ki te 4\sqrt{3}.

$$y=\frac{4\sqrt{3}-4}{-8}$$

Whakawehe -4+4\sqrt{3} ki te -8.

$$y=\frac{1-\sqrt{3}}{2}$$

Nā, me whakaoti te whārite y=\frac{-4±4\sqrt{3}}{-8} ina he tango te ±. Tango 4\sqrt{3} mai i -4.

$$y=\frac{-4\sqrt{3}-4}{-8}$$

Whakawehe -4-4\sqrt{3} ki te -8.

$$y=\frac{\sqrt{3}+1}{2}$$

Kua oti te whārite te whakatau.

$$y=\frac{1-\sqrt{3}}{2} y=\frac{\sqrt{3}+1}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te y ki te 1-3y.

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

Whakamahia te āhuatanga tohatoha hei whakarea te y ki te y-3.

$$2+y-3y^{2}=y^{2}-3y$$

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

$$2+y-3y^{2}-y^{2}=-3y$$

Pahekotia te -3y^{2} me -y^{2}, ka -4y^{2}.

$$2+y-4y^{2}=-3y$$

Me tāpiri te 3y ki ngā taha e rua.

$$2+y-4y^{2}+3y=0$$

Pahekotia te y me 3y, ka 4y.

$$2+4y-4y^{2}=0$$

Tangohia te 2 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$4y-4y^{2}=-2$$

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.

$$-4y^{2}+4y=-2$$

Whakawehea ngā taha e rua ki te -4.

$$\frac{-4y^{2}+4y}{-4}=-\frac{2}{-4}$$

Mā te whakawehe ki te -4 ka wetekia te whakareanga ki te -4.

$$y^{2}+\frac{4}{-4}y=-\frac{2}{-4}$$

Whakawehe 4 ki te -4.

$$y^{2}-y=-\frac{2}{-4}$$

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

$$y^{2}-y=\frac{1}{2}$$

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

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

Pūruatia -\frac{1}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$y^{2}-y+\frac{1}{4}=\frac{1}{2}+\frac{1}{4}$$

Tāpiri \frac{1}{2} ki te \frac{1}{4} 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.

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

Tauwehea y^{2}-y+\frac{1}{4}. 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}.

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

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

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

Whakarūnātia.

$$y-\frac{1}{2}=\frac{\sqrt{3}}{2} y-\frac{1}{2}=-\frac{\sqrt{3}}{2}$$

Me tāpiri \frac{1}{2} ki ngā taha e rua o te whārite.

$$y=\frac{\sqrt{3}+1}{2} y=\frac{1-\sqrt{3}}{2}$$