Whakaoti i te x^2-7x+12=0

$${ x }^{ 2 } -7x+12 = 0$$

Whakaoti mō x

$$x=3$$

$$x=4$$

Ngā Upane Whakamahi Tauwehe

Hei whakaoti i te whārite, whakatauwehea te x^{2}-7x+12 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.

$$a+b=-7 ab=12$$

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 12.

$$-1,-12 -2,-6 -3,-4$$

Tātaihia te tapeke mō ia takirua.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

Ko te otinga te takirua ka hoatu i te tapeke -7.

$$a=-4 b=-3$$

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

$$\left(x-4\right)\left(x-3\right)$$

Hei kimi otinga whārite, me whakaoti te x-4=0 me te x-3=0.

$$x=4 x=3$$

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+12. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-7 ab=1\times 12=12$$

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 12.

$$-1,-12 -2,-6 -3,-4$$

Tātaihia te tapeke mō ia takirua.

$$-1-12=-13 -2-6=-8 -3-4=-7$$

Ko te otinga te takirua ka hoatu i te tapeke -7.

$$a=-4 b=-3$$

Tuhia anō te x^{2}-7x+12 hei \left(x^{2}-4x\right)+\left(-3x+12\right).

$$\left(x^{2}-4x\right)+\left(-3x+12\right)$$

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

$$x\left(x-4\right)-3\left(x-4\right)$$

Whakatauwehea atu te kīanga pātahi x-4 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-4\right)\left(x-3\right)$$

Hei kimi otinga whārite, me whakaoti te x-4=0 me te x-3=0.

$$x=4 x=3$$

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.

$$x^{2}-7x+12=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -7 mō b, me 12 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 12}}{2}$$

Pūrua -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 12}}{2}$$

Whakareatia -4 ki te 12.

$$x=\frac{-\left(-7\right)±\sqrt{49-48}}{2}$$

Tāpiri 49 ki te -48.

$$x=\frac{-\left(-7\right)±\sqrt{1}}{2}$$

Tuhia te pūtakerua o te 1.

$$x=\frac{-\left(-7\right)±1}{2}$$

Ko te tauaro o -7 ko 7.

$$x=\frac{7±1}{2}$$

Nā, me whakaoti te whārite x=\frac{7±1}{2} ina he tāpiri te ±. Tāpiri 7 ki te 1.

$$x=\frac{8}{2}$$

Whakawehe 8 ki te 2.

$$x=4$$

Nā, me whakaoti te whārite x=\frac{7±1}{2} ina he tango te ±. Tango 1 mai i 7.

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

Whakawehe 6 ki te 2.

$$x=3$$

Kua oti te whārite te whakatau.

$$x=4 x=3$$

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.

$$x^{2}-7x+12=0$$

Me tango 12 mai i ngā taha e rua o te whārite.

$$x^{2}-7x+12-12=-12$$

Mā te tango i te 12 i a ia ake anō ka toe ko te 0.

$$x^{2}-7x=-12$$

Whakawehea te -7, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{7}{2}. Nā, tāpiria te pūrua o te -\frac{7}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=-12+\left(-\frac{7}{2}\right)^{2}$$

Pūruatia -\frac{7}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-7x+\frac{49}{4}=-12+\frac{49}{4}$$

Tāpiri -12 ki te \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=\frac{1}{4}$$

Tauwehea x^{2}-7x+\frac{49}{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}.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{1}{4}$$

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

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Whakarūnātia.

$$x-\frac{7}{2}=\frac{1}{2} x-\frac{7}{2}=-\frac{1}{2}$$

Me tāpiri \frac{7}{2} ki ngā taha e rua o te whārite.

$$x=4 x=3$$