Whakaoti i te k^2-k=8

$$k ^ { 2 } - k = 8$$

Whakaoti mō k

$$k = \frac{\sqrt{33} + 1}{2} \approx 3.372281323$$

$$k=\frac{1-\sqrt{33}}{2}\approx -2.372281323$$

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.

$$k^{2}-k=8$$

Me tango 8 mai i ngā taha e rua o te whārite.

$$k^{2}-k-8=8-8$$

Mā te tango i te 8 i a ia ake anō ka toe ko te 0.

$$k^{2}-k-8=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -1 mō b, me -8 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\left(-1\right)±\sqrt{1-4\left(-8\right)}}{2}$$

Whakareatia -4 ki te -8.

$$k=\frac{-\left(-1\right)±\sqrt{1+32}}{2}$$

Tāpiri 1 ki te 32.

$$k=\frac{-\left(-1\right)±\sqrt{33}}{2}$$

Ko te tauaro o -1 ko 1.

$$k=\frac{1±\sqrt{33}}{2}$$

Nā, me whakaoti te whārite k=\frac{1±\sqrt{33}}{2} ina he tāpiri te ±. Tāpiri 1 ki te \sqrt{33}.

$$k=\frac{\sqrt{33}+1}{2}$$

Nā, me whakaoti te whārite k=\frac{1±\sqrt{33}}{2} ina he tango te ±. Tango \sqrt{33} mai i 1.

$$k=\frac{1-\sqrt{33}}{2}$$

Kua oti te whārite te whakatau.

$$k=\frac{\sqrt{33}+1}{2} k=\frac{1-\sqrt{33}}{2}$$

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.

$$k^{2}-k=8$$

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.

$$k^{2}-k+\left(-\frac{1}{2}\right)^{2}=8+\left(-\frac{1}{2}\right)^{2}$$

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

$$k^{2}-k+\frac{1}{4}=8+\frac{1}{4}$$

Tāpiri 8 ki te \frac{1}{4}.

$$k^{2}-k+\frac{1}{4}=\frac{33}{4}$$

Tauwehea te k^{2}-k+\frac{1}{4}. Ko te tikanga, ina ko x^{2}+bx+c he pūrua tika, ka taea te tauwehe i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(k-\frac{1}{2}\right)^{2}=\frac{33}{4}$$

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

$$\sqrt{\left(k-\frac{1}{2}\right)^{2}}=\sqrt{\frac{33}{4}}$$

Whakarūnātia.

$$k-\frac{1}{2}=\frac{\sqrt{33}}{2} k-\frac{1}{2}=-\frac{\sqrt{33}}{2}$$

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

$$k=\frac{\sqrt{33}+1}{2} k=\frac{1-\sqrt{33}}{2}$$