Whakaoti i te x^2+2x+3=12

$${ x }^{ 2 } +2x+3=12$$

Whakaoti mō x (complex solution)

$$x=\sqrt{10}-1\approx 2.16227766$$

$$x=-\left(\sqrt{10}+1\right)\approx -4.16227766$$

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.

$$x^{2}+2x+3=12$$

Me tango 12 mai i ngā taha e rua o te whārite.

$$x^{2}+2x+3-12=12-12$$

Mā te tango i te 12 i a ia ake anō ka toe ko te 0.

$$x^{2}+2x+3-12=0$$

Tango 12 mai i 3.

$$x^{2}+2x-9=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 2 mō b, me -9 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-9\right)}}{2}$$

Pūrua 2.

$$x=\frac{-2±\sqrt{4-4\left(-9\right)}}{2}$$

Whakareatia -4 ki te -9.

$$x=\frac{-2±\sqrt{4+36}}{2}$$

Tāpiri 4 ki te 36.

$$x=\frac{-2±\sqrt{40}}{2}$$

Tuhia te pūtakerua o te 40.

$$x=\frac{-2±2\sqrt{10}}{2}$$

Nā, me whakaoti te whārite x=\frac{-2±2\sqrt{10}}{2} ina he tāpiri te ±. Tāpiri -2 ki te 2\sqrt{10}.

$$x=\frac{2\sqrt{10}-2}{2}$$

Whakawehe -2+2\sqrt{10} ki te 2.

$$x=\sqrt{10}-1$$

Nā, me whakaoti te whārite x=\frac{-2±2\sqrt{10}}{2} ina he tango te ±. Tango 2\sqrt{10} mai i -2.

$$x=\frac{-2\sqrt{10}-2}{2}$$

Whakawehe -2-2\sqrt{10} ki te 2.

$$x=-\sqrt{10}-1$$

Kua oti te whārite te whakatau.

$$x=\sqrt{10}-1 x=-\sqrt{10}-1$$

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.

$$x^{2}+2x+3=12$$

Me tango 3 mai i ngā taha e rua o te whārite.

$$x^{2}+2x+3-3=12-3$$

Mā te tango i te 3 i a ia ake anō ka toe ko te 0.

$$x^{2}+2x=12-3$$

Tango 3 mai i 12.

$$x^{2}+2x=9$$

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

$$x^{2}+2x+1^{2}=9+1^{2}$$

Pūrua 1.

$$x^{2}+2x+1=9+1$$

Tāpiri 9 ki te 1.

$$x^{2}+2x+1=10$$

Tauwehea te x^{2}+2x+1. 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}.

$$\left(x+1\right)^{2}=10$$

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

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{10}$$

Whakarūnātia.

$$x+1=\sqrt{10} x+1=-\sqrt{10}$$

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

$$x=\sqrt{10}-1 x=-\sqrt{10}-1$$

Whakaoti mō x

$$x=\sqrt{10}-1\approx 2.16227766$$

$$x=-\sqrt{10}-1\approx -4.16227766$$

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.

$$x^{2}+2x+3=12$$

Me tango 12 mai i ngā taha e rua o te whārite.

$$x^{2}+2x+3-12=12-12$$

Mā te tango i te 12 i a ia ake anō ka toe ko te 0.

$$x^{2}+2x+3-12=0$$

Tango 12 mai i 3.

$$x^{2}+2x-9=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 2 mō b, me -9 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-9\right)}}{2}$$

Pūrua 2.

$$x=\frac{-2±\sqrt{4-4\left(-9\right)}}{2}$$

Whakareatia -4 ki te -9.

$$x=\frac{-2±\sqrt{4+36}}{2}$$

Tāpiri 4 ki te 36.

$$x=\frac{-2±\sqrt{40}}{2}$$

Tuhia te pūtakerua o te 40.

$$x=\frac{-2±2\sqrt{10}}{2}$$

Nā, me whakaoti te whārite x=\frac{-2±2\sqrt{10}}{2} ina he tāpiri te ±. Tāpiri -2 ki te 2\sqrt{10}.

$$x=\frac{2\sqrt{10}-2}{2}$$

Whakawehe -2+2\sqrt{10} ki te 2.

$$x=\sqrt{10}-1$$

Nā, me whakaoti te whārite x=\frac{-2±2\sqrt{10}}{2} ina he tango te ±. Tango 2\sqrt{10} mai i -2.

$$x=\frac{-2\sqrt{10}-2}{2}$$

Whakawehe -2-2\sqrt{10} ki te 2.

$$x=-\sqrt{10}-1$$

Kua oti te whārite te whakatau.

$$x=\sqrt{10}-1 x=-\sqrt{10}-1$$

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.

$$x^{2}+2x+3=12$$

Me tango 3 mai i ngā taha e rua o te whārite.

$$x^{2}+2x+3-3=12-3$$

Mā te tango i te 3 i a ia ake anō ka toe ko te 0.

$$x^{2}+2x=12-3$$

Tango 3 mai i 12.

$$x^{2}+2x=9$$

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

$$x^{2}+2x+1^{2}=9+1^{2}$$

Pūrua 1.

$$x^{2}+2x+1=9+1$$

Tāpiri 9 ki te 1.

$$x^{2}+2x+1=10$$

Tauwehea te x^{2}+2x+1. 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}.

$$\left(x+1\right)^{2}=10$$

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

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{10}$$

Whakarūnātia.

$$x+1=\sqrt{10} x+1=-\sqrt{10}$$

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

$$x=\sqrt{10}-1 x=-\sqrt{10}-1$$