Whakaoti i te n^2-sqrt{3}n+1=0

$$n ^ { 2 } - \sqrt { 3 } n + 1 = 0$$

Whakaoti mō n

$$n=\frac{\sqrt{3}+i}{2}\approx 0.866025404+0.5i$$

$$n=\frac{\sqrt{3}-i}{2}\approx 0.866025404-0.5i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakaraupapatia anō ngā kīanga tau.

$$n^{2}-\sqrt{3}n+1=0$$

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.

$$n^{2}+\left(-\sqrt{3}\right)n+1=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -\sqrt{3} mō b, me 1 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Pūrua -\sqrt{3}.

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

Tāpiri 3 ki te -4.

$$n=\frac{-\left(-\sqrt{3}\right)±\sqrt{-1}}{2}$$

Tuhia te pūtakerua o te -1.

$$n=\frac{-\left(-\sqrt{3}\right)±i}{2}$$

Ko te tauaro o -\sqrt{3} ko \sqrt{3}.

$$n=\frac{\sqrt{3}±i}{2}$$

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

$$n=\frac{\sqrt{3}+i}{2}$$

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

$$n=\frac{\sqrt{3}}{2}+\frac{1}{2}i$$

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

$$n=\frac{\sqrt{3}-i}{2}$$

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

$$n=\frac{\sqrt{3}}{2}-\frac{1}{2}i$$

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 1 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$n^{2}-\sqrt{3}n=-1$$

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.

$$n^{2}+\left(-\sqrt{3}\right)n=-1$$

Whakawehea te -\sqrt{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{\sqrt{3}}{2}. Nā, tāpiria te pūrua o te -\frac{\sqrt{3}}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūrua -\frac{\sqrt{3}}{2}.

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

Tāpiri -1 ki te \frac{3}{4}.

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

Tauwehea n^{2}+\left(-\sqrt{3}\right)n+\frac{3}{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}.

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

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

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

Whakarūnātia.

$$n-\frac{\sqrt{3}}{2}=\frac{1}{2}i n-\frac{\sqrt{3}}{2}=-\frac{1}{2}i$$

Me tāpiri \frac{\sqrt{3}}{2} ki ngā taha e rua o te whārite.

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