Whakaoti i te x^2-3x+2=0

$${ x }^{ 2 } -3x+2 = 0$$

Whakaoti mō x

$$x=1$$

$$x=2$$

Ngā Upane Whakamahi Tauwehe

Hei whakaoti i te whārite, whakatauwehea te x^{2}-3x+2 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.

$$a+b=-3 ab=2$$

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. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=-2 b=-1$$

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

$$\left(x-2\right)\left(x-1\right)$$

Hei kimi otinga whārite, me whakaoti te x-2=0 me te x-1=0.

$$x=2 x=1$$

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+2. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-3 ab=1\times 2=2$$

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. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=-2 b=-1$$

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

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

Tauwehea te x i te tuatahi me te -1 i te rōpū tuarua.

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

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

$$\left(x-2\right)\left(x-1\right)$$

Hei kimi otinga whārite, me whakaoti te x-2=0 me te x-1=0.

$$x=2 x=1$$

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.

$$x^{2}-3x+2=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -3 mō b, me 2 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2}}{2}$$

Pūrua -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 2}}{2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-3\right)±\sqrt{9-8}}{2}$$

Tāpiri 9 ki te -8.

$$x=\frac{-\left(-3\right)±\sqrt{1}}{2}$$

Tuhia te pūtakerua o te 1.

$$x=\frac{-\left(-3\right)±1}{2}$$

Ko te tauaro o -3 ko 3.

$$x=\frac{3±1}{2}$$

Nā, me whakaoti te whārite x=\frac{3±1}{2} ina he tāpiri te ±. Tāpiri 3 ki te 1.

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

Whakawehe 4 ki te 2.

$$x=2$$

Nā, me whakaoti te whārite x=\frac{3±1}{2} ina he tango te ±. Tango 1 mai i 3.

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

Whakawehe 2 ki te 2.

$$x=1$$

Kua oti te whārite te whakatau.

$$x=2 x=1$$

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.

$$x^{2}-3x+2=0$$

Me tango 2 mai i ngā taha e rua o te whārite.

$$x^{2}-3x+2-2=-2$$

Mā te tango i te 2 i a ia ake anō ka toe ko te 0.

$$x^{2}-3x=-2$$

Whakawehea te -3, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{3}{2}. Nā, tāpiria te pūrua o te -\frac{3}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=-2+\left(-\frac{3}{2}\right)^{2}$$

Pūruatia -\frac{3}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-3x+\frac{9}{4}=-2+\frac{9}{4}$$

Tāpiri -2 ki te \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{1}{4}$$

Tauwehea te x^{2}-3x+\frac{9}{4}. 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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{1}{4}$$

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

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Whakarūnātia.

$$x-\frac{3}{2}=\frac{1}{2} x-\frac{3}{2}=-\frac{1}{2}$$

Me tāpiri \frac{3}{2} ki ngā taha e rua o te whārite.

$$x=2 x=1$$