Whakaoti i te p^2-4+p=0

$$p ^ { 2 } - 4 + p = 0$$

Whakaoti mō p

$$p = \frac{\sqrt{17} - 1}{2} \approx 1.561552813$$

$$p=\frac{-\sqrt{17}-1}{2}\approx -2.561552813$$

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.

$$p^{2}+p-4=0$$

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

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

Pūrua 1.

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

Whakareatia -4 ki te -4.

$$p=\frac{-1±\sqrt{1+16}}{2}$$

Tāpiri 1 ki te 16.

$$p=\frac{-1±\sqrt{17}}{2}$$

Nā, me whakaoti te whārite p=\frac{-1±\sqrt{17}}{2} ina he tāpiri te ±. Tāpiri -1 ki te \sqrt{17}.

$$p=\frac{\sqrt{17}-1}{2}$$

Nā, me whakaoti te whārite p=\frac{-1±\sqrt{17}}{2} ina he tango te ±. Tango \sqrt{17} mai i -1.

$$p=\frac{-\sqrt{17}-1}{2}$$

Kua oti te whārite te whakatau.

$$p=\frac{\sqrt{17}-1}{2} p=\frac{-\sqrt{17}-1}{2}$$

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.

$$p^{2}+p-4=0$$

Me tāpiri 4 ki ngā taha e rua o te whārite.

$$p^{2}+p-4-\left(-4\right)=-\left(-4\right)$$

Mā te tango i te -4 i a ia ake anō ka toe ko te 0.

$$p^{2}+p=-\left(-4\right)$$

Tango -4 mai i 0.

$$p^{2}+p=4$$

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.

$$p^{2}+p+\left(\frac{1}{2}\right)^{2}=4+\left(\frac{1}{2}\right)^{2}$$

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

$$p^{2}+p+\frac{1}{4}=4+\frac{1}{4}$$

Tāpiri 4 ki te \frac{1}{4}.

$$p^{2}+p+\frac{1}{4}=\frac{17}{4}$$

Tauwehea p^{2}+p+\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}.

$$\left(p+\frac{1}{2}\right)^{2}=\frac{17}{4}$$

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

$$\sqrt{\left(p+\frac{1}{2}\right)^{2}}=\sqrt{\frac{17}{4}}$$

Whakarūnātia.

$$p+\frac{1}{2}=\frac{\sqrt{17}}{2} p+\frac{1}{2}=-\frac{\sqrt{17}}{2}$$

Me tango \frac{1}{2} mai i ngā taha e rua o te whārite.

$$p=\frac{\sqrt{17}-1}{2} p=\frac{-\sqrt{17}-1}{2}$$