Whakaoti i te 2*(a^2-9)+a=15

$$2 \cdot ( a ^ { 2 } - 9 ) + a = 15$$

Whakaoti mō a

$$a = \frac{\sqrt{265} - 1}{4} \approx 3.819705149$$

$$a=\frac{-\sqrt{265}-1}{4}\approx -4.319705149$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te a^{2}-9.

$$2a^{2}-18+a=15$$

Tangohia te 15 mai i ngā taha e rua.

$$2a^{2}-18+a-15=0$$

Tangohia te 15 i te -18, ka -33.

$$2a^{2}-33+a=0$$

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.

$$2a^{2}+a-33=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 2 mō a, 1 mō b, me -33 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Pūrua 1.

$$a=\frac{-1±\sqrt{1-4\times 2\left(-33\right)}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$a=\frac{-1±\sqrt{1-8\left(-33\right)}}{2\times 2}$$

Whakareatia -8 ki te -33.

$$a=\frac{-1±\sqrt{1+264}}{2\times 2}$$

Tāpiri 1 ki te 264.

$$a=\frac{-1±\sqrt{265}}{2\times 2}$$

Whakareatia 2 ki te 2.

$$a=\frac{-1±\sqrt{265}}{4}$$

Nā, me whakaoti te whārite a=\frac{-1±\sqrt{265}}{4} ina he tāpiri te ±. Tāpiri -1 ki te \sqrt{265}.

$$a=\frac{\sqrt{265}-1}{4}$$

Nā, me whakaoti te whārite a=\frac{-1±\sqrt{265}}{4} ina he tango te ±. Tango \sqrt{265} mai i -1.

$$a=\frac{-\sqrt{265}-1}{4}$$

Kua oti te whārite te whakatau.

$$a=\frac{\sqrt{265}-1}{4} a=\frac{-\sqrt{265}-1}{4}$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te a^{2}-9.

$$2a^{2}-18+a=15$$

Me tāpiri te 18 ki ngā taha e rua.

$$2a^{2}+a=15+18$$

Tāpirihia te 15 ki te 18, ka 33.

$$2a^{2}+a=33$$

Whakawehea ngā taha e rua ki te 2.

$$\frac{2a^{2}+a}{2}=\frac{33}{2}$$

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

$$a^{2}+\frac{1}{2}a=\frac{33}{2}$$

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.

$$a^{2}+\frac{1}{2}a+\left(\frac{1}{4}\right)^{2}=\frac{33}{2}+\left(\frac{1}{4}\right)^{2}$$

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

$$a^{2}+\frac{1}{2}a+\frac{1}{16}=\frac{33}{2}+\frac{1}{16}$$

Tāpiri \frac{33}{2} ki te \frac{1}{16} mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$a^{2}+\frac{1}{2}a+\frac{1}{16}=\frac{265}{16}$$

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}.

$$\left(a+\frac{1}{4}\right)^{2}=\frac{265}{16}$$

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

$$\sqrt{\left(a+\frac{1}{4}\right)^{2}}=\sqrt{\frac{265}{16}}$$

Whakarūnātia.

$$a+\frac{1}{4}=\frac{\sqrt{265}}{4} a+\frac{1}{4}=-\frac{\sqrt{265}}{4}$$

Me tango \frac{1}{4} mai i ngā taha e rua o te whārite.

$$a=\frac{\sqrt{265}-1}{4} a=\frac{-\sqrt{265}-1}{4}$$