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

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

Whakaoti mō x

$$x=-1$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Whakawehea ngā taha e rua ki te 2.

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

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei x^{2}+ax+bx+1. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=2 ab=1\times 1=1$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōrunga te a+b, he tōrunga hoki a a me b. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=1 b=1$$

Tuhia anō te x^{2}+2x+1 hei \left(x^{2}+x\right)+\left(x+1\right).

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

Whakatauwehea atu x i te x^{2}+x.

$$x\left(x+1\right)+x+1$$

Whakatauwehea atu te kīanga pātahi x+1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x+1\right)\left(x+1\right)$$

Tuhia anōtia hei pūrua huarua.

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

Hei kimi i te otinga whārite, whakaotia te x+1=0.

$$x=-1$$

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.

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

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

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

Pūrua 4.

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

Whakareatia -4 ki te 2.

$$x=\frac{-4±\sqrt{16-8\times 2}}{2\times 2}$$

Whakareatia -8 ki te 2.

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

Tāpiri 16 ki te -16.

$$x=\frac{-4±\sqrt{0}}{2\times 2}$$

Tuhia te pūtakerua o te 0.

$$x=-\frac{4}{2\times 2}$$

Whakareatia 2 ki te 2.

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

Whakawehe -4 ki te 4.

$$x=-1$$

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.

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

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

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

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

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

Whakawehea ngā taha e rua ki te 2.

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

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

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

Whakawehe 4 ki te 2.

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

Whakawehe -2 ki te 2.

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

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.

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

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}.

$$\left(x+1\right)^{2}=0$$

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

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{0}$$

Whakarūnātia.

$$x+1=0 x+1=0$$

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

$$x=-1 x=-1$$

Kua oti te whārite te whakatau. He ōrite ngā whakatau.

$$x=-1$$