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

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

Whakaoti mō x

$$x=2$$

Ngā Upane Whakamahi Tauwehe

Hei whakaoti i te whārite, whakatauwehea te x^{2}-4x+4 mā te whakamahi i te tātai x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-4 ab=4$$

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. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 4.

$$-1,-4 -2,-2$$

Tātaihia te tapeke mō ia takirua.

$$-1-4=-5 -2-2=-4$$

Ko te otinga te takirua ka hoatu i te tapeke -4.

$$a=-2 b=-2$$

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

$$\left(x-2\right)\left(x-2\right)$$

Tuhia anōtia hei pūrua huarua.

$$\left(x-2\right)^{2}$$

Hei kimi i te otinga whārite, whakaotia te x-2=0.

$$x=2$$

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 x^{2}+ax+bx+4. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-4 ab=1\times 4=4$$

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. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 4.

$$-1,-4 -2,-2$$

Tātaihia te tapeke mō ia takirua.

$$-1-4=-5 -2-2=-4$$

Ko te otinga te takirua ka hoatu i te tapeke -4.

$$a=-2 b=-2$$

Tuhia anō te x^{2}-4x+4 hei \left(x^{2}-2x\right)+\left(-2x+4\right).

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

Tauwehea te x i te tuatahi me te -2 i te rōpū tuarua.

$$x\left(x-2\right)-2\left(x-2\right)$$

Whakatauwehea atu te kīanga pātahi x-2 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-2\right)\left(x-2\right)$$

Tuhia anōtia hei pūrua huarua.

$$\left(x-2\right)^{2}$$

Hei kimi i te otinga whārite, whakaotia te x-2=0.

$$x=2$$

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.

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

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

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

Pūrua -4.

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

Whakareatia -4 ki te 4.

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

Tāpiri 16 ki te -16.

$$x=\frac{-\left(-4\right)±\sqrt{0}}{2}$$

Tuhia te pūtakerua o te 0.

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

Ko te tauaro o -4 ko 4.

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

Whakawehe 4 ki te 2.

$$x=2$$

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.

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

Tauwehea x^{2}-4x+4. 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}.

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

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

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

Whakarūnātia.

$$x-2=0 x-2=0$$

Me tāpiri 2 ki ngā taha e rua o te whārite.

$$x=2 x=2$$

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

$$x=2$$