Whakaoti i te x^2+2x=1

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

Whakaoti mō x (complex solution)

$$x=\sqrt{2}-1\approx 0.414213562$$

$$x=-\left(\sqrt{2}+1\right)\approx -2.414213562$$

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.

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

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

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

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

$$x^{2}+2x-1=0$$

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

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

Pūrua 2.

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

Whakareatia -4 ki te -1.

$$x=\frac{-2±\sqrt{4+4}}{2}$$

Tāpiri 4 ki te 4.

$$x=\frac{-2±\sqrt{8}}{2}$$

Tuhia te pūtakerua o te 8.

$$x=\frac{-2±2\sqrt{2}}{2}$$

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

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

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

$$x=\sqrt{2}-1$$

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

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

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

$$x=-\sqrt{2}-1$$

Kua oti te whārite te whakatau.

$$x=\sqrt{2}-1 x=-\sqrt{2}-1$$

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.

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

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.

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

Pūrua 1.

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

Tāpiri 1 ki te 1.

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

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

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

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

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

Whakarūnātia.

$$x+1=\sqrt{2} x+1=-\sqrt{2}$$

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

$$x=\sqrt{2}-1 x=-\sqrt{2}-1$$

Whakaoti mō x

$$x=\sqrt{2}-1\approx 0.414213562$$

$$x=-\sqrt{2}-1\approx -2.414213562$$

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.

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

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

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

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

$$x^{2}+2x-1=0$$

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

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

Pūrua 2.

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

Whakareatia -4 ki te -1.

$$x=\frac{-2±\sqrt{4+4}}{2}$$

Tāpiri 4 ki te 4.

$$x=\frac{-2±\sqrt{8}}{2}$$

Tuhia te pūtakerua o te 8.

$$x=\frac{-2±2\sqrt{2}}{2}$$

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

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

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

$$x=\sqrt{2}-1$$

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

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

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

$$x=-\sqrt{2}-1$$

Kua oti te whārite te whakatau.

$$x=\sqrt{2}-1 x=-\sqrt{2}-1$$

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.

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

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.

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

Pūrua 1.

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

Tāpiri 1 ki te 1.

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

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

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

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

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

Whakarūnātia.

$$x+1=\sqrt{2} x+1=-\sqrt{2}$$

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

$$x=\sqrt{2}-1 x=-\sqrt{2}-1$$