Whakaoti i te s^2+3s+1=0

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

Whakaoti mō s

$$s=\frac{\sqrt{5}-3}{2}\approx -0.381966011$$

$$s=\frac{-\sqrt{5}-3}{2}\approx -2.618033989$$

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.

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

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}.

$$s=\frac{-3±\sqrt{3^{2}-4}}{2}$$

Pūrua 3.

$$s=\frac{-3±\sqrt{9-4}}{2}$$

Tāpiri 9 ki te -4.

$$s=\frac{-3±\sqrt{5}}{2}$$

Nā, me whakaoti te whārite s=\frac{-3±\sqrt{5}}{2} ina he tāpiri te ±. Tāpiri -3 ki te \sqrt{5}.

$$s=\frac{\sqrt{5}-3}{2}$$

Nā, me whakaoti te whārite s=\frac{-3±\sqrt{5}}{2} ina he tango te ±. Tango \sqrt{5} mai i -3.

$$s=\frac{-\sqrt{5}-3}{2}$$

Kua oti te whārite te whakatau.

$$s=\frac{\sqrt{5}-3}{2} s=\frac{-\sqrt{5}-3}{2}$$

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.

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

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.

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

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}.

$$s^{2}+3s+\frac{9}{4}=\frac{5}{4}$$

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}.

$$\left(s+\frac{3}{2}\right)^{2}=\frac{5}{4}$$

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

$$\sqrt{\left(s+\frac{3}{2}\right)^{2}}=\sqrt{\frac{5}{4}}$$

Whakarūnātia.

$$s+\frac{3}{2}=\frac{\sqrt{5}}{2} s+\frac{3}{2}=-\frac{\sqrt{5}}{2}$$

Me tango \frac{3}{2} mai i ngā taha e rua o te whārite.

$$s=\frac{\sqrt{5}-3}{2} s=\frac{-\sqrt{5}-3}{2}$$