Whakaoti i te 2y^2+2y-1=0

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

Whakaoti mō y

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

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

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.

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

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

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

Pūrua 2.

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

Whakareatia -4 ki te 2.

$$y=\frac{-2±\sqrt{4-8\left(-1\right)}}{2\times 2}$$

Whakareatia -8 ki te -1.

$$y=\frac{-2±\sqrt{4+8}}{2\times 2}$$

Tāpiri 4 ki te 8.

$$y=\frac{-2±\sqrt{12}}{2\times 2}$$

Tuhia te pūtakerua o te 12.

$$y=\frac{-2±2\sqrt{3}}{2\times 2}$$

Whakareatia 2 ki te 2.

$$y=\frac{-2±2\sqrt{3}}{4}$$

Nā, me whakaoti te whārite y=\frac{-2±2\sqrt{3}}{4} ina he tāpiri te ±. Tāpiri -2 ki te 2\sqrt{3}.

$$y=\frac{2\sqrt{3}-2}{4}$$

Whakawehe -2+2\sqrt{3} ki te 4.

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

Nā, me whakaoti te whārite y=\frac{-2±2\sqrt{3}}{4} ina he tango te ±. Tango 2\sqrt{3} mai i -2.

$$y=\frac{-2\sqrt{3}-2}{4}$$

Whakawehe -2-2\sqrt{3} ki te 4.

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

Kua oti te whārite te whakatau.

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

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.

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

Me tāpiri 1 ki ngā taha e rua o te whārite.

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

Mā te tango i te -1 i a ia ake anō ka toe ko te 0.

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

Tango -1 mai i 0.

$$2y^{2}+2y=1$$

Whakawehea ngā taha e rua ki te 2.

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

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

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

Whakawehe 2 ki 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 tango \frac{1}{2} mai i ngā taha e rua o te whārite.

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