Whakaoti i te 2a^2-a-2=0

$$2 a ^ { 2 } - a - 2 = 0$$

Whakaoti mō a

$$a = \frac{\sqrt{17} + 1}{4} \approx 1.280776406$$

$$a=\frac{1-\sqrt{17}}{4}\approx -0.780776406$$

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.

$$2a^{2}-a-2=0$$

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

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

Whakareatia -4 ki te 2.

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

Whakareatia -8 ki te -2.

$$a=\frac{-\left(-1\right)±\sqrt{1+16}}{2\times 2}$$

Tāpiri 1 ki te 16.

$$a=\frac{-\left(-1\right)±\sqrt{17}}{2\times 2}$$

Ko te tauaro o -1 ko 1.

$$a=\frac{1±\sqrt{17}}{2\times 2}$$

Whakareatia 2 ki te 2.

$$a=\frac{1±\sqrt{17}}{4}$$

Nā, me whakaoti te whārite a=\frac{1±\sqrt{17}}{4} ina he tāpiri te ±. Tāpiri 1 ki te \sqrt{17}.

$$a=\frac{\sqrt{17}+1}{4}$$

Nā, me whakaoti te whārite a=\frac{1±\sqrt{17}}{4} ina he tango te ±. Tango \sqrt{17} mai i 1.

$$a=\frac{1-\sqrt{17}}{4}$$

Kua oti te whārite te whakatau.

$$a=\frac{\sqrt{17}+1}{4} a=\frac{1-\sqrt{17}}{4}$$

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.

$$2a^{2}-a-2=0$$

Me tāpiri 2 ki ngā taha e rua o te whārite.

$$2a^{2}-a-2-\left(-2\right)=-\left(-2\right)$$

Mā te tango i te -2 i a ia ake anō ka toe ko te 0.

$$2a^{2}-a=-\left(-2\right)$$

Tango -2 mai i 0.

$$2a^{2}-a=2$$

Whakawehea ngā taha e rua ki te 2.

$$\frac{2a^{2}-a}{2}=\frac{2}{2}$$

Mā te whakawehe ki te 2 ka wetekia te whakareanga ki te 2.

$$a^{2}-\frac{1}{2}a=\frac{2}{2}$$

Whakawehe 2 ki te 2.

$$a^{2}-\frac{1}{2}a=1$$

Whakawehea te -\frac{1}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{4}. Nā, tāpiria te pūrua o te -\frac{1}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia -\frac{1}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$a^{2}-\frac{1}{2}a+\frac{1}{16}=1+\frac{1}{16}$$

Tāpiri 1 ki te \frac{1}{16}.

$$a^{2}-\frac{1}{2}a+\frac{1}{16}=\frac{17}{16}$$

Tauwehea a^{2}-\frac{1}{2}a+\frac{1}{16}. 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}.

$$\left(a-\frac{1}{4}\right)^{2}=\frac{17}{16}$$

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

$$\sqrt{\left(a-\frac{1}{4}\right)^{2}}=\sqrt{\frac{17}{16}}$$

Whakarūnātia.

$$a-\frac{1}{4}=\frac{\sqrt{17}}{4} a-\frac{1}{4}=-\frac{\sqrt{17}}{4}$$

Me tāpiri \frac{1}{4} ki ngā taha e rua o te whārite.

$$a=\frac{\sqrt{17}+1}{4} a=\frac{1-\sqrt{17}}{4}$$