Whakaoti i te t^2-2t+1=0

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

Whakaoti mō t

$$t=1$$

Ngā Upane Whakamahi Tauwehe

Hei whakaoti i te whārite, whakatauwehea te t^{2}-2t+1 mā te whakamahi i te tātai t^{2}+\left(a+b\right)t+ab=\left(t+a\right)\left(t+b\right). Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-2 ab=1$$

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

$$a=-1 b=-1$$

Me tuhi anō te kīanga whakatauwehe \left(t+a\right)\left(t+b\right) mā ngā uara i tātaihia.

$$\left(t-1\right)\left(t-1\right)$$

Tuhia anōtia hei pūrua huarua.

$$\left(t-1\right)^{2}$$

Hei kimi i te otinga whārite, whakaotia te t-1=0.

$$t=1$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei t^{2}+at+bt+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 tōraro te a+b, he tōraro hoki a a me b. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=-1 b=-1$$

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

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

Tauwehea te t i te tuatahi me te -1 i te rōpū tuarua.

$$t\left(t-1\right)-\left(t-1\right)$$

Whakatauwehea atu te kīanga pātahi t-1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(t-1\right)\left(t-1\right)$$

Tuhia anōtia hei pūrua huarua.

$$\left(t-1\right)^{2}$$

Hei kimi i te otinga whārite, whakaotia te t-1=0.

$$t=1$$

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.

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

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}.

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

Pūrua -2.

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

Tāpiri 4 ki te -4.

$$t=\frac{-\left(-2\right)±\sqrt{0}}{2}$$

Tuhia te pūtakerua o te 0.

$$t=-\frac{-2}{2}$$

Ko te tauaro o -2 ko 2.

$$t=\frac{2}{2}$$

Whakawehe 2 ki te 2.

$$t=1$$

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.

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

Tauwehea t^{2}-2t+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}.

$$\left(t-1\right)^{2}=0$$

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

$$\sqrt{\left(t-1\right)^{2}}=\sqrt{0}$$

Whakarūnātia.

$$t-1=0 t-1=0$$

Me tāpiri 1 ki ngā taha e rua o te whārite.

$$t=1 t=1$$

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

$$t=1$$