Whakaoti i te 1/2x^2+x=2

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

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.

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

Me tango 2 mai i ngā taha e rua o te whārite.

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

Mā te tango i te 2 i a ia ake anō ka toe ko te 0.

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

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

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

Pūrua 1.

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

Whakareatia -4 ki te \frac{1}{2}.

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

Whakareatia -2 ki te -2.

$$x=\frac{-1±\sqrt{1+4}}{2\times \frac{1}{2}}$$

Tāpiri 1 ki te 4.

$$x=\frac{-1±\sqrt{5}}{2\times \frac{1}{2}}$$

Whakareatia 2 ki te \frac{1}{2}.

$$x=\frac{-1±\sqrt{5}}{1}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{5}}{1} ina he tāpiri te ±. Tāpiri -1 ki te \sqrt{5}.

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

Whakawehe -1+\sqrt{5} ki te 1.

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

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{5}}{1} ina he tango te ±. Tango \sqrt{5} mai i -1.

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

Whakawehe -1-\sqrt{5} ki te 1.

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

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.

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

Me whakarea ngā taha e rua ki te 2.

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

Mā te whakawehe ki te \frac{1}{2} ka wetekia te whakareanga ki te \frac{1}{2}.

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

Whakawehe 1 ki te \frac{1}{2} mā te whakarea 1 ki te tau huripoki o \frac{1}{2}.

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

Whakawehe 2 ki te \frac{1}{2} mā te whakarea 2 ki te tau huripoki o \frac{1}{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.

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

Me tango 2 mai i ngā taha e rua o te whārite.

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

Mā te tango i te 2 i a ia ake anō ka toe ko te 0.

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

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

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

Pūrua 1.

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

Whakareatia -4 ki te \frac{1}{2}.

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

Whakareatia -2 ki te -2.

$$x=\frac{-1±\sqrt{1+4}}{2\times \frac{1}{2}}$$

Tāpiri 1 ki te 4.

$$x=\frac{-1±\sqrt{5}}{2\times \frac{1}{2}}$$

Whakareatia 2 ki te \frac{1}{2}.

$$x=\frac{-1±\sqrt{5}}{1}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{5}}{1} ina he tāpiri te ±. Tāpiri -1 ki te \sqrt{5}.

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

Whakawehe -1+\sqrt{5} ki te 1.

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

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{5}}{1} ina he tango te ±. Tango \sqrt{5} mai i -1.

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

Whakawehe -1-\sqrt{5} ki te 1.

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

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.

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

Me whakarea ngā taha e rua ki te 2.

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

Mā te whakawehe ki te \frac{1}{2} ka wetekia te whakareanga ki te \frac{1}{2}.

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

Whakawehe 1 ki te \frac{1}{2} mā te whakarea 1 ki te tau huripoki o \frac{1}{2}.

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

Whakawehe 2 ki te \frac{1}{2} mā te whakarea 2 ki te tau huripoki o \frac{1}{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.

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