Whakaoti i te x^2-10x+25=0

$$x^2-10x+25=0$$

Whakaoti mō x

$$x=5$$

Ngā Upane Whakamahi Tauwehe

Hei whakaoti i te whārite, whakatauwehea te x^{2}-10x+25 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.

$$a+b=-10 ab=25$$

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 25.

$$-1,-25 -5,-5$$

Tātaihia te tapeke mō ia takirua.

$$-1-25=-26 -5-5=-10$$

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

$$a=-5 b=-5$$

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

$$\left(x-5\right)\left(x-5\right)$$

Tuhia anōtia hei pūrua huarua.

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

Hei kimi i te otinga whārite, whakaotia te x-5=0.

$$x=5$$

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+25. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-10 ab=1\times 25=25$$

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 25.

$$-1,-25 -5,-5$$

Tātaihia te tapeke mō ia takirua.

$$-1-25=-26 -5-5=-10$$

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

$$a=-5 b=-5$$

Tuhia anō te x^{2}-10x+25 hei \left(x^{2}-5x\right)+\left(-5x+25\right).

$$\left(x^{2}-5x\right)+\left(-5x+25\right)$$

Tauwehea te x i te tuatahi me te -5 i te rōpū tuarua.

$$x\left(x-5\right)-5\left(x-5\right)$$

Whakatauwehea atu te kīanga pātahi x-5 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-5\right)\left(x-5\right)$$

Tuhia anōtia hei pūrua huarua.

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

Hei kimi i te otinga whārite, whakaotia te x-5=0.

$$x=5$$

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.

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

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -10 mō b, me 25 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 25}}{2}$$

Pūrua -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 25}}{2}$$

Whakareatia -4 ki te 25.

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

Tāpiri 100 ki te -100.

$$x=\frac{-\left(-10\right)±\sqrt{0}}{2}$$

Tuhia te pūtakerua o te 0.

$$x=-\frac{-10}{2}$$

Ko te tauaro o -10 ko 10.

$$x=\frac{10}{2}$$

Whakawehe 10 ki te 2.

$$x=5$$

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.

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

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}.

$$\left(x-5\right)^{2}=0$$

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

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{0}$$

Whakarūnātia.

$$x-5=0 x-5=0$$

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

$$x=5 x=5$$

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

$$x=5$$