Whakaoti i te s^2-3s=1

$$s ^ { 2 } - 3 s = 1$$

Whakaoti mō s

$$s = \frac{\sqrt{13} + 3}{2} \approx 3.302775638$$

$$s=\frac{3-\sqrt{13}}{2}\approx -0.302775638$$

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.

$$s^{2}-3s=1$$

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

$$s^{2}-3s-1=1-1$$

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

$$s^{2}-3s-1=0$$

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

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

Pūrua -3.

$$s=\frac{-\left(-3\right)±\sqrt{9-4\left(-1\right)}}{2}$$

Whakareatia -4 ki te -1.

$$s=\frac{-\left(-3\right)±\sqrt{9+4}}{2}$$

Tāpiri 9 ki te 4.

$$s=\frac{-\left(-3\right)±\sqrt{13}}{2}$$

Ko te tauaro o -3 ko 3.

$$s=\frac{3±\sqrt{13}}{2}$$

Nā, me whakaoti te whārite s=\frac{3±\sqrt{13}}{2} ina he tāpiri te ±. Tāpiri 3 ki te \sqrt{13}.

$$s=\frac{\sqrt{13}+3}{2}$$

Nā, me whakaoti te whārite s=\frac{3±\sqrt{13}}{2} ina he tango te ±. Tango \sqrt{13} mai i 3.

$$s=\frac{3-\sqrt{13}}{2}$$

Kua oti te whārite te whakatau.

$$s=\frac{\sqrt{13}+3}{2} s=\frac{3-\sqrt{13}}{2}$$

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.

$$s^{2}-3s=1$$

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

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

Pūruatia -\frac{3}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$s^{2}-3s+\frac{9}{4}=1+\frac{9}{4}$$

Tāpiri 1 ki te \frac{9}{4}.

$$s^{2}-3s+\frac{9}{4}=\frac{13}{4}$$

Tauwehea s^{2}-3s+\frac{9}{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}.

$$\left(s-\frac{3}{2}\right)^{2}=\frac{13}{4}$$

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

$$\sqrt{\left(s-\frac{3}{2}\right)^{2}}=\sqrt{\frac{13}{4}}$$

Whakarūnātia.

$$s-\frac{3}{2}=\frac{\sqrt{13}}{2} s-\frac{3}{2}=-\frac{\sqrt{13}}{2}$$

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

$$s=\frac{\sqrt{13}+3}{2} s=\frac{3-\sqrt{13}}{2}$$