Whakaoti i te (x-5)^2-9=0

$$( x - 5 ) ^ { 2 } - 9 = 0$$

Whakaoti mō x

$$x=8$$

$$x=2$$

Ngā Upane Whakamahi Tauwehe

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x-5\right)^{2}.

$$x^{2}-10x+25-9=0$$

Tangohia te 9 i te 25, ka 16.

$$x^{2}-10x+16=0$$

Hei whakaoti i te whārite, whakatauwehea te x^{2}-10x+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.

$$a+b=-10 ab=16$$

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 -10.

$$a=-8 b=-2$$

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

$$\left(x-8\right)\left(x-2\right)$$

Hei kimi otinga whārite, me whakaoti te x-8=0 me te x-2=0.

$$x=8 x=2$$

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-5\right)^{2}.

$$x^{2}-10x+25-9=0$$

Tangohia te 9 i te 25, ka 16.

$$x^{2}-10x+16=0$$

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.

$$a+b=-10 ab=1\times 16=16$$

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 -10.

$$a=-8 b=-2$$

Tuhia anō te x^{2}-10x+16 hei \left(x^{2}-8x\right)+\left(-2x+16\right).

$$\left(x^{2}-8x\right)+\left(-2x+16\right)$$

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

$$x\left(x-8\right)-2\left(x-8\right)$$

Whakatauwehea atu te kīanga pātahi x-8 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-8\right)\left(x-2\right)$$

Hei kimi otinga whārite, me whakaoti te x-8=0 me te x-2=0.

$$x=8 x=2$$

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-5\right)^{2}.

$$x^{2}-10x+25-9=0$$

Tangohia te 9 i te 25, ka 16.

$$x^{2}-10x+16=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -10 mō b, me 16 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 16}}{2}$$

Pūrua -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 16}}{2}$$

Whakareatia -4 ki te 16.

$$x=\frac{-\left(-10\right)±\sqrt{100-64}}{2}$$

Tāpiri 100 ki te -64.

$$x=\frac{-\left(-10\right)±\sqrt{36}}{2}$$

Tuhia te pūtakerua o te 36.

$$x=\frac{-\left(-10\right)±6}{2}$$

Ko te tauaro o -10 ko 10.

$$x=\frac{10±6}{2}$$

Nā, me whakaoti te whārite x=\frac{10±6}{2} ina he tāpiri te ±. Tāpiri 10 ki te 6.

$$x=\frac{16}{2}$$

Whakawehe 16 ki te 2.

$$x=8$$

Nā, me whakaoti te whārite x=\frac{10±6}{2} ina he tango te ±. Tango 6 mai i 10.

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

Whakawehe 4 ki te 2.

$$x=2$$

Kua oti te whārite te whakatau.

$$x=8 x=2$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(x-5\right)^{2}.

$$x^{2}-10x+25-9=0$$

Tangohia te 9 i te 25, ka 16.

$$x^{2}-10x+16=0$$

Tangohia te 16 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$x^{2}-10x=-16$$

Whakawehea te -10, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -5. Nā, tāpiria te pūrua o te -5 ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-10x+\left(-5\right)^{2}=-16+\left(-5\right)^{2}$$

Pūrua -5.

$$x^{2}-10x+25=-16+25$$

Tāpiri -16 ki te 25.

$$x^{2}-10x+25=9$$

Tauwehea te x^{2}-10x+25. Ko te tikanga, ina ko x^{2}+bx+c he pūrua tika, ka taea te tauwehe i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=9$$

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

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{9}$$

Whakarūnātia.

$$x-5=3 x-5=-3$$

Me tāpiri 5 ki ngā taha e rua o te whārite.

$$x=8 x=2$$