Whakaoti i te x^2+2x+4=8

$${ x }^{ 2 } +2x+4=8$$

Whakaoti mō x (complex solution)

$$x=\sqrt{5}-1\approx 1.236067977$$

$$x=-\left(\sqrt{5}+1\right)\approx -3.236067977$$

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.

$$x^{2}+2x+4=8$$

Me tango 8 mai i ngā taha e rua o te whārite.

$$x^{2}+2x+4-8=8-8$$

Mā te tango i te 8 i a ia ake anō ka toe ko te 0.

$$x^{2}+2x+4-8=0$$

Tango 8 mai i 4.

$$x^{2}+2x-4=0$$

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

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

Pūrua 2.

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

Whakareatia -4 ki te -4.

$$x=\frac{-2±\sqrt{4+16}}{2}$$

Tāpiri 4 ki te 16.

$$x=\frac{-2±\sqrt{20}}{2}$$

Tuhia te pūtakerua o te 20.

$$x=\frac{-2±2\sqrt{5}}{2}$$

Nā, me whakaoti te whārite x=\frac{-2±2\sqrt{5}}{2} ina he tāpiri te ±. Tāpiri -2 ki te 2\sqrt{5}.

$$x=\frac{2\sqrt{5}-2}{2}$$

Whakawehe -2+2\sqrt{5} ki te 2.

$$x=\sqrt{5}-1$$

Nā, me whakaoti te whārite x=\frac{-2±2\sqrt{5}}{2} ina he tango te ±. Tango 2\sqrt{5} mai i -2.

$$x=\frac{-2\sqrt{5}-2}{2}$$

Whakawehe -2-2\sqrt{5} ki te 2.

$$x=-\sqrt{5}-1$$

Kua oti te whārite te whakatau.

$$x=\sqrt{5}-1 x=-\sqrt{5}-1$$

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.

$$x^{2}+2x+4=8$$

Me tango 4 mai i ngā taha e rua o te whārite.

$$x^{2}+2x+4-4=8-4$$

Mā te tango i te 4 i a ia ake anō ka toe ko te 0.

$$x^{2}+2x=8-4$$

Tango 4 mai i 8.

$$x^{2}+2x=4$$

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.

$$x^{2}+2x+1^{2}=4+1^{2}$$

Pūrua 1.

$$x^{2}+2x+1=4+1$$

Tāpiri 4 ki te 1.

$$x^{2}+2x+1=5$$

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}.

$$\left(x+1\right)^{2}=5$$

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

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{5}$$

Whakarūnātia.

$$x+1=\sqrt{5} x+1=-\sqrt{5}$$

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

$$x=\sqrt{5}-1 x=-\sqrt{5}-1$$

Whakaoti mō x

$$x=\sqrt{5}-1\approx 1.236067977$$

$$x=-\sqrt{5}-1\approx -3.236067977$$

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.

$$x^{2}+2x+4=8$$

Me tango 8 mai i ngā taha e rua o te whārite.

$$x^{2}+2x+4-8=8-8$$

Mā te tango i te 8 i a ia ake anō ka toe ko te 0.

$$x^{2}+2x+4-8=0$$

Tango 8 mai i 4.

$$x^{2}+2x-4=0$$

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

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

Pūrua 2.

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

Whakareatia -4 ki te -4.

$$x=\frac{-2±\sqrt{4+16}}{2}$$

Tāpiri 4 ki te 16.

$$x=\frac{-2±\sqrt{20}}{2}$$

Tuhia te pūtakerua o te 20.

$$x=\frac{-2±2\sqrt{5}}{2}$$

Nā, me whakaoti te whārite x=\frac{-2±2\sqrt{5}}{2} ina he tāpiri te ±. Tāpiri -2 ki te 2\sqrt{5}.

$$x=\frac{2\sqrt{5}-2}{2}$$

Whakawehe -2+2\sqrt{5} ki te 2.

$$x=\sqrt{5}-1$$

Nā, me whakaoti te whārite x=\frac{-2±2\sqrt{5}}{2} ina he tango te ±. Tango 2\sqrt{5} mai i -2.

$$x=\frac{-2\sqrt{5}-2}{2}$$

Whakawehe -2-2\sqrt{5} ki te 2.

$$x=-\sqrt{5}-1$$

Kua oti te whārite te whakatau.

$$x=\sqrt{5}-1 x=-\sqrt{5}-1$$

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.

$$x^{2}+2x+4=8$$

Me tango 4 mai i ngā taha e rua o te whārite.

$$x^{2}+2x+4-4=8-4$$

Mā te tango i te 4 i a ia ake anō ka toe ko te 0.

$$x^{2}+2x=8-4$$

Tango 4 mai i 8.

$$x^{2}+2x=4$$

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.

$$x^{2}+2x+1^{2}=4+1^{2}$$

Pūrua 1.

$$x^{2}+2x+1=4+1$$

Tāpiri 4 ki te 1.

$$x^{2}+2x+1=5$$

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}.

$$\left(x+1\right)^{2}=5$$

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

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{5}$$

Whakarūnātia.

$$x+1=\sqrt{5} x+1=-\sqrt{5}$$

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

$$x=\sqrt{5}-1 x=-\sqrt{5}-1$$