Whakaoti i te (x-5)^2-64=0

$$( x - 5 ) ^ { 2 } - 64 = 0$$

Whakaoti mō x

$$x=13$$

$$x=-3$$

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}.

$$x^{2}-10x+25-64=0$$

Tangohia te 64 i te 25, ka -39.

$$x^{2}-10x-39=0$$

Hei whakaoti i te whārite, whakatauwehea te x^{2}-10x-39 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.

$$a+b=-10 ab=-39$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -39.

$$1,-39 3,-13$$

Tātaihia te tapeke mō ia takirua.

$$1-39=-38 3-13=-10$$

Ko te otinga te takirua ka hoatu i te tapeke -10.

$$a=-13 b=3$$

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

$$\left(x-13\right)\left(x+3\right)$$

Hei kimi otinga whārite, me whakaoti te x-13=0 me te x+3=0.

$$x=13 x=-3$$

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}.

$$x^{2}-10x+25-64=0$$

Tangohia te 64 i te 25, ka -39.

$$x^{2}-10x-39=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei x^{2}+ax+bx-39. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-10 ab=1\left(-39\right)=-39$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -39.

$$1,-39 3,-13$$

Tātaihia te tapeke mō ia takirua.

$$1-39=-38 3-13=-10$$

Ko te otinga te takirua ka hoatu i te tapeke -10.

$$a=-13 b=3$$

Tuhia anō te x^{2}-10x-39 hei \left(x^{2}-13x\right)+\left(3x-39\right).

$$\left(x^{2}-13x\right)+\left(3x-39\right)$$

Tauwehea te x i te tuatahi me te 3 i te rōpū tuarua.

$$x\left(x-13\right)+3\left(x-13\right)$$

Whakatauwehea atu te kīanga pātahi x-13 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-13\right)\left(x+3\right)$$

Hei kimi otinga whārite, me whakaoti te x-13=0 me te x+3=0.

$$x=13 x=-3$$

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}.

$$x^{2}-10x+25-64=0$$

Tangohia te 64 i te 25, ka -39.

$$x^{2}-10x-39=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -10 mō b, me -39 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-39\right)}}{2}$$

Pūrua -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-39\right)}}{2}$$

Whakareatia -4 ki te -39.

$$x=\frac{-\left(-10\right)±\sqrt{100+156}}{2}$$

Tāpiri 100 ki te 156.

$$x=\frac{-\left(-10\right)±\sqrt{256}}{2}$$

Tuhia te pūtakerua o te 256.

$$x=\frac{-\left(-10\right)±16}{2}$$

Ko te tauaro o -10 ko 10.

$$x=\frac{10±16}{2}$$

Nā, me whakaoti te whārite x=\frac{10±16}{2} ina he tāpiri te ±. Tāpiri 10 ki te 16.

$$x=\frac{26}{2}$$

Whakawehe 26 ki te 2.

$$x=13$$

Nā, me whakaoti te whārite x=\frac{10±16}{2} ina he tango te ±. Tango 16 mai i 10.

$$x=-\frac{6}{2}$$

Whakawehe -6 ki te 2.

$$x=-3$$

Kua oti te whārite te whakatau.

$$x=13 x=-3$$

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}.

$$x^{2}-10x+25-64=0$$

Tangohia te 64 i te 25, ka -39.

$$x^{2}-10x-39=0$$

Me tāpiri te 39 ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

$$x^{2}-10x=39$$

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.

$$x^{2}-10x+\left(-5\right)^{2}=39+\left(-5\right)^{2}$$

Pūrua -5.

$$x^{2}-10x+25=39+25$$

Tāpiri 39 ki te 25.

$$x^{2}-10x+25=64$$

Tauwehea x^{2}-10x+25. 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}.

$$\left(x-5\right)^{2}=64$$

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

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{64}$$

Whakarūnātia.

$$x-5=8 x-5=-8$$

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

$$x=13 x=-3$$