Whakaoti i te left(1xright)^2-5x+6=0

$${ \left(1x \right) }^{ 2 } -5x+6 = 0$$

Whakaoti mō x

$$x=2$$

$$x=3$$

Ngā Upane Whakamahi Tauwehe

Whakarohaina te \left(1x\right)^{2}.

$$1^{2}x^{2}-5x+6=0$$

Tātaihia te 1 mā te pū o 2, kia riro ko 1.

$$1x^{2}-5x+6=0$$

Whakaraupapatia anō ngā kīanga tau.

$$x^{2}-5x+6=0$$

Hei whakaoti i te whārite, whakatauwehea te x^{2}-5x+6 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.

$$a+b=-5 ab=6$$

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 6.

$$-1,-6 -2,-3$$

Tātaihia te tapeke mō ia takirua.

$$-1-6=-7 -2-3=-5$$

Ko te otinga te takirua ka hoatu i te tapeke -5.

$$a=-3 b=-2$$

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

$$\left(x-3\right)\left(x-2\right)$$

Hei kimi otinga whārite, me whakaoti te x-3=0 me te x-2=0.

$$x=3 x=2$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Whakarohaina te \left(1x\right)^{2}.

$$1^{2}x^{2}-5x+6=0$$

Tātaihia te 1 mā te pū o 2, kia riro ko 1.

$$1x^{2}-5x+6=0$$

Whakaraupapatia anō ngā kīanga tau.

$$x^{2}-5x+6=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei x^{2}+ax+bx+6. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-5 ab=1\times 6=6$$

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 6.

$$-1,-6 -2,-3$$

Tātaihia te tapeke mō ia takirua.

$$-1-6=-7 -2-3=-5$$

Ko te otinga te takirua ka hoatu i te tapeke -5.

$$a=-3 b=-2$$

Tuhia anō te x^{2}-5x+6 hei \left(x^{2}-3x\right)+\left(-2x+6\right).

$$\left(x^{2}-3x\right)+\left(-2x+6\right)$$

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

$$x\left(x-3\right)-2\left(x-3\right)$$

Whakatauwehea atu te kīanga pātahi x-3 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-3\right)\left(x-2\right)$$

Hei kimi otinga whārite, me whakaoti te x-3=0 me te x-2=0.

$$x=3 x=2$$

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.

$$x^{2}-5x+6=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -5 mō b, me 6 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 6}}{2}$$

Pūrua -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 6}}{2}$$

Whakareatia -4 ki te 6.

$$x=\frac{-\left(-5\right)±\sqrt{25-24}}{2}$$

Tāpiri 25 ki te -24.

$$x=\frac{-\left(-5\right)±\sqrt{1}}{2}$$

Tuhia te pūtakerua o te 1.

$$x=\frac{-\left(-5\right)±1}{2}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±1}{2}$$

Nā, me whakaoti te whārite x=\frac{5±1}{2} ina he tāpiri te ±. Tāpiri 5 ki te 1.

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

Whakawehe 6 ki te 2.

$$x=3$$

Nā, me whakaoti te whārite x=\frac{5±1}{2} ina he tango te ±. Tango 1 mai i 5.

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

Whakawehe 4 ki te 2.

$$x=2$$

Kua oti te whārite te whakatau.

$$x=3 x=2$$

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.

$$x^{2}-5x+6=0$$

Me tango 6 mai i ngā taha e rua o te whārite.

$$x^{2}-5x+6-6=-6$$

Mā te tango i te 6 i a ia ake anō ka toe ko te 0.

$$x^{2}-5x=-6$$

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

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-6+\left(-\frac{5}{2}\right)^{2}$$

Pūruatia -\frac{5}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-5x+\frac{25}{4}=-6+\frac{25}{4}$$

Tāpiri -6 ki te \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{1}{4}$$

Tauwehea x^{2}-5x+\frac{25}{4}. 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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{1}{4}$$

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

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Whakarūnātia.

$$x-\frac{5}{2}=\frac{1}{2} x-\frac{5}{2}=-\frac{1}{2}$$

Me tāpiri \frac{5}{2} ki ngā taha e rua o te whārite.

$$x=3 x=2$$