Whakaoti i te 3a+(a^2+1)=1

$$3 a + ( a ^ { 2 } + 1 ) = 1$$

Whakaoti mō a

$$a=-3$$

$$a=0$$

Ngā Upane Whakamahi Tauwehe

Tangohia te 1 mai i ngā taha e rua.

$$3a+a^{2}+1-1=0$$

Tangohia te 1 i te 1, ka 0.

$$3a+a^{2}=0$$

Tauwehea te a.

$$a\left(3+a\right)=0$$

Hei kimi otinga whārite, me whakaoti te a=0 me te 3+a=0.

$$a=0 a=-3$$

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.

$$a^{2}+3a+1=1$$

Me tango 1 mai i ngā taha e rua o te whārite.

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

Mā te tango i te 1 i a ia ake anō ka toe ko te 0.

$$a^{2}+3a+1-1=0$$

Tango 1 mai i 1.

$$a^{2}+3a=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 3 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-3±\sqrt{3^{2}}}{2}$$

Tuhia te pūtakerua o te 3^{2}.

$$a=\frac{-3±3}{2}$$

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

$$a=\frac{0}{2}$$

Whakawehe 0 ki te 2.

$$a=0$$

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

$$a=-\frac{6}{2}$$

Whakawehe -6 ki te 2.

$$a=-3$$

Kua oti te whārite te whakatau.

$$a=0 a=-3$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 1 mai i ngā taha e rua.

$$3a+a^{2}+1-1=0$$

Tangohia te 1 i te 1, ka 0.

$$3a+a^{2}=0$$

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.

$$a^{2}+3a=0$$

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.

$$a^{2}+3a+\left(\frac{3}{2}\right)^{2}=\left(\frac{3}{2}\right)^{2}$$

Pūruatia \frac{3}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

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

Tauwehea a^{2}+3a+\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}.

$$\left(a+\frac{3}{2}\right)^{2}=\frac{9}{4}$$

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

$$\sqrt{\left(a+\frac{3}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Whakarūnātia.

$$a+\frac{3}{2}=\frac{3}{2} a+\frac{3}{2}=-\frac{3}{2}$$

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

$$a=0 a=-3$$