Whakaoti i te (x-4)^2=0

$$( x - 4 ) ^ { 2 } = 0$$

Whakaoti mō x

$$x=4$$

Ngā Upane Whakamahi Tauwehe

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x-4\right)^{2}.

$$x^{2}-8x+16=0$$

Hei whakaoti i te whārite, whakatauwehea te x^{2}-8x+16 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.

$$a+b=-8 ab=16$$

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 16.

$$-1,-16 -2,-8 -4,-4$$

Tātaihia te tapeke mō ia takirua.

$$-1-16=-17 -2-8=-10 -4-4=-8$$

Ko te otinga te takirua ka hoatu i te tapeke -8.

$$a=-4 b=-4$$

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

$$\left(x-4\right)\left(x-4\right)$$

Tuhia anōtia hei pūrua huarua.

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

Hei kimi i te otinga whārite, whakaotia te x-4=0.

$$x=4$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x-4\right)^{2}.

$$x^{2}-8x+16=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei x^{2}+ax+bx+16. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-8 ab=1\times 16=16$$

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 16.

$$-1,-16 -2,-8 -4,-4$$

Tātaihia te tapeke mō ia takirua.

$$-1-16=-17 -2-8=-10 -4-4=-8$$

Ko te otinga te takirua ka hoatu i te tapeke -8.

$$a=-4 b=-4$$

Tuhia anō te x^{2}-8x+16 hei \left(x^{2}-4x\right)+\left(-4x+16\right).

$$\left(x^{2}-4x\right)+\left(-4x+16\right)$$

Tauwehea te x i te tuatahi me te -4 i te rōpū tuarua.

$$x\left(x-4\right)-4\left(x-4\right)$$

Whakatauwehea atu te kīanga pātahi x-4 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-4\right)\left(x-4\right)$$

Tuhia anōtia hei pūrua huarua.

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

Hei kimi i te otinga whārite, whakaotia te x-4=0.

$$x=4$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x-4\right)^{2}.

$$x^{2}-8x+16=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -8 mō b, me 16 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Pūrua -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 16}}{2}$$

Whakareatia -4 ki te 16.

$$x=\frac{-\left(-8\right)±\sqrt{64-64}}{2}$$

Tāpiri 64 ki te -64.

$$x=\frac{-\left(-8\right)±\sqrt{0}}{2}$$

Tuhia te pūtakerua o te 0.

$$x=-\frac{-8}{2}$$

Ko te tauaro o -8 ko 8.

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

Whakawehe 8 ki te 2.

$$x=4$$

Ngā Upane mō te Whakaoti i te Pūrua

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

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

Whakarūnātia.

$$x-4=0 x-4=0$$

Me tāpiri 4 ki ngā taha e rua o te whārite.

$$x=4 x=4$$

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

$$x=4$$