Whakaoti i te 4x+2x^2-8=0

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

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.

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

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 2 mō a, 4 mō b, me -8 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 2\left(-8\right)}}{2\times 2}$$

Pūrua 4.

$$x=\frac{-4±\sqrt{16-4\times 2\left(-8\right)}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-4±\sqrt{16-8\left(-8\right)}}{2\times 2}$$

Whakareatia -8 ki te -8.

$$x=\frac{-4±\sqrt{16+64}}{2\times 2}$$

Tāpiri 16 ki te 64.

$$x=\frac{-4±\sqrt{80}}{2\times 2}$$

Tuhia te pūtakerua o te 80.

$$x=\frac{-4±4\sqrt{5}}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{-4±4\sqrt{5}}{4}$$

Nā, me whakaoti te whārite x=\frac{-4±4\sqrt{5}}{4} ina he tāpiri te ±. Tāpiri -4 ki te 4\sqrt{5}.

$$x=\frac{4\sqrt{5}-4}{4}$$

Whakawehe -4+4\sqrt{5} ki te 4.

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

Nā, me whakaoti te whārite x=\frac{-4±4\sqrt{5}}{4} ina he tango te ±. Tango 4\sqrt{5} mai i -4.

$$x=\frac{-4\sqrt{5}-4}{4}$$

Whakawehe -4-4\sqrt{5} ki te 4.

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

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.

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

Me tāpiri 8 ki ngā taha e rua o te whārite.

$$2x^{2}+4x-8-\left(-8\right)=-\left(-8\right)$$

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

$$2x^{2}+4x=-\left(-8\right)$$

Tango -8 mai i 0.

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

Whakawehea ngā taha e rua ki te 2.

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

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

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

Whakawehe 4 ki te 2.

$$x^{2}+2x=\frac{8}{2}$$

Whakawehe 8 ki te 2.

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

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.

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

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 2 mō a, 4 mō b, me -8 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-4±\sqrt{4^{2}-4\times 2\left(-8\right)}}{2\times 2}$$

Pūrua 4.

$$x=\frac{-4±\sqrt{16-4\times 2\left(-8\right)}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-4±\sqrt{16-8\left(-8\right)}}{2\times 2}$$

Whakareatia -8 ki te -8.

$$x=\frac{-4±\sqrt{16+64}}{2\times 2}$$

Tāpiri 16 ki te 64.

$$x=\frac{-4±\sqrt{80}}{2\times 2}$$

Tuhia te pūtakerua o te 80.

$$x=\frac{-4±4\sqrt{5}}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{-4±4\sqrt{5}}{4}$$

Nā, me whakaoti te whārite x=\frac{-4±4\sqrt{5}}{4} ina he tāpiri te ±. Tāpiri -4 ki te 4\sqrt{5}.

$$x=\frac{4\sqrt{5}-4}{4}$$

Whakawehe -4+4\sqrt{5} ki te 4.

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

Nā, me whakaoti te whārite x=\frac{-4±4\sqrt{5}}{4} ina he tango te ±. Tango 4\sqrt{5} mai i -4.

$$x=\frac{-4\sqrt{5}-4}{4}$$

Whakawehe -4-4\sqrt{5} ki te 4.

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

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.

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

Me tāpiri 8 ki ngā taha e rua o te whārite.

$$2x^{2}+4x-8-\left(-8\right)=-\left(-8\right)$$

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

$$2x^{2}+4x=-\left(-8\right)$$

Tango -8 mai i 0.

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

Whakawehea ngā taha e rua ki te 2.

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

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

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

Whakawehe 4 ki te 2.

$$x^{2}+2x=\frac{8}{2}$$

Whakawehe 8 ki te 2.

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

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.

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