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

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

Whakaoti mō x

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

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

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.

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

Me tango 1 mai i ngā taha e rua o te whārite.

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

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

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

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}.

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

Pūrua -2.

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

Whakareatia -4 ki te 2.

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

Whakareatia -8 ki te -1.

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

Tāpiri 4 ki te 8.

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

Tuhia te pūtakerua o te 12.

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

Ko te tauaro o -2 ko 2.

$$x=\frac{2±2\sqrt{3}}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{2±2\sqrt{3}}{4}$$

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

$$x=\frac{2\sqrt{3}+2}{4}$$

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

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

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

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

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

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

Kua oti te whārite te whakatau.

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

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.

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

Whakawehea ngā taha e rua ki te 2.

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

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

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

Whakawehe -2 ki te 2.

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

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.

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

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

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

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.

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

Tauwehea x^{2}-x+\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}.

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

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

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

Whakarūnātia.

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

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

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