Whakaoti i te x^2-x-1=0

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

Whakaoti mō x

$$x = \frac{\sqrt{5} + 1}{2} \approx 1.618033989$$

$$x=\frac{1-\sqrt{5}}{2}\approx -0.618033989$$

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.

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

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

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

Whakareatia -4 ki te -1.

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

Tāpiri 1 ki te 4.

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

Ko te tauaro o -1 ko 1.

$$x=\frac{1±\sqrt{5}}{2}$$

Nā, me whakaoti te whārite x=\frac{1±\sqrt{5}}{2} ina he tāpiri te ±. Tāpiri 1 ki te \sqrt{5}.

$$x=\frac{\sqrt{5}+1}{2}$$

Nā, me whakaoti te whārite x=\frac{1±\sqrt{5}}{2} ina he tango te ±. Tango \sqrt{5} mai i 1.

$$x=\frac{1-\sqrt{5}}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{5}+1}{2} x=\frac{1-\sqrt{5}}{2}$$

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.

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

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

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

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

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

Tango -1 mai i 0.

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

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.

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

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

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

Tāpiri 1 ki te \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{5}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{5}{4}$$

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

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{5}{4}}$$

Whakarūnātia.

$$x-\frac{1}{2}=\frac{\sqrt{5}}{2} x-\frac{1}{2}=-\frac{\sqrt{5}}{2}$$

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

$$x=\frac{\sqrt{5}+1}{2} x=\frac{1-\sqrt{5}}{2}$$