Whakaoti i te 2y^2-y+2=0

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

Whakaoti mō y

$$y=\frac{1+\sqrt{15}i}{4}\approx 0.25+0.968245837i$$

$$y=\frac{-\sqrt{15}i+1}{4}\approx 0.25-0.968245837i$$

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.

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

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

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

Whakareatia -4 ki te 2.

$$y=\frac{-\left(-1\right)±\sqrt{1-8\times 2}}{2\times 2}$$

Whakareatia -8 ki te 2.

$$y=\frac{-\left(-1\right)±\sqrt{1-16}}{2\times 2}$$

Tāpiri 1 ki te -16.

$$y=\frac{-\left(-1\right)±\sqrt{-15}}{2\times 2}$$

Tuhia te pūtakerua o te -15.

$$y=\frac{-\left(-1\right)±\sqrt{15}i}{2\times 2}$$

Ko te tauaro o -1 ko 1.

$$y=\frac{1±\sqrt{15}i}{2\times 2}$$

Whakareatia 2 ki te 2.

$$y=\frac{1±\sqrt{15}i}{4}$$

Nā, me whakaoti te whārite y=\frac{1±\sqrt{15}i}{4} ina he tāpiri te ±. Tāpiri 1 ki te i\sqrt{15}.

$$y=\frac{1+\sqrt{15}i}{4}$$

Nā, me whakaoti te whārite y=\frac{1±\sqrt{15}i}{4} ina he tango te ±. Tango i\sqrt{15} mai i 1.

$$y=\frac{-\sqrt{15}i+1}{4}$$

Kua oti te whārite te whakatau.

$$y=\frac{1+\sqrt{15}i}{4} y=\frac{-\sqrt{15}i+1}{4}$$

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.

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

Me tango 2 mai i ngā taha e rua o te whārite.

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

Mā te tango i te 2 i a ia ake anō ka toe ko te 0.

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

Whakawehea ngā taha e rua ki te 2.

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

Mā te whakawehe ki te 2 ka wetekia te whakareanga ki te 2.

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

Whakawehe -2 ki te 2.

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

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.

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

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

$$y^{2}-\frac{1}{2}y+\frac{1}{16}=-1+\frac{1}{16}$$

Tāpiri -1 ki te \frac{1}{16}.

$$y^{2}-\frac{1}{2}y+\frac{1}{16}=-\frac{15}{16}$$

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

$$\left(y-\frac{1}{4}\right)^{2}=-\frac{15}{16}$$

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

$$\sqrt{\left(y-\frac{1}{4}\right)^{2}}=\sqrt{-\frac{15}{16}}$$

Whakarūnātia.

$$y-\frac{1}{4}=\frac{\sqrt{15}i}{4} y-\frac{1}{4}=-\frac{\sqrt{15}i}{4}$$

Me tāpiri \frac{1}{4} ki ngā taha e rua o te whārite.

$$y=\frac{1+\sqrt{15}i}{4} y=\frac{-\sqrt{15}i+1}{4}$$