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

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

Whakaoti mō x

$$x=-4$$

$$x=3$$

Ngā Upane Whakamahi Tauwehe

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

$$a+b=1 ab=-12$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōrunga te a+b, he nui ake te uara pū o te tau tōrunga i tō te tōraro. 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.

$$-1+12=11 -2+6=4 -3+4=1$$

Ko te otinga te takirua ka hoatu i te tapeke 1.

$$a=-3 b=4$$

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

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

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

$$x=3 x=-4$$

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.

$$a+b=1 ab=1\left(-12\right)=-12$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōrunga te a+b, he nui ake te uara pū o te tau tōrunga i tō te tōraro. 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.

$$-1+12=11 -2+6=4 -3+4=1$$

Ko te otinga te takirua ka hoatu i te tapeke 1.

$$a=-3 b=4$$

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

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

Tauwehea te x i te tuatahi me te 4 i te rōpū tuarua.

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

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

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

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

$$x=3 x=-4$$

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.

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

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

$$x=\frac{-1±\sqrt{1^{2}-4\left(-12\right)}}{2}$$

Pūrua 1.

$$x=\frac{-1±\sqrt{1-4\left(-12\right)}}{2}$$

Whakareatia -4 ki te -12.

$$x=\frac{-1±\sqrt{1+48}}{2}$$

Tāpiri 1 ki te 48.

$$x=\frac{-1±\sqrt{49}}{2}$$

Tuhia te pūtakerua o te 49.

$$x=\frac{-1±7}{2}$$

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

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

Whakawehe 6 ki te 2.

$$x=3$$

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

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

Whakawehe -8 ki te 2.

$$x=-4$$

Kua oti te whārite te whakatau.

$$x=3 x=-4$$

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.

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

Me tāpiri 12 ki ngā taha e rua o te whārite.

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

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

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

Tango -12 mai i 0.

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

Whakawehea te 1, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{2}. Nā, tāpiria te pūrua o te \frac{1}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=12+\left(\frac{1}{2}\right)^{2}$$

Pūruatia \frac{1}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+x+\frac{1}{4}=12+\frac{1}{4}$$

Tāpiri 12 ki te \frac{1}{4}.

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

Tauwehea x^{2}+x+\frac{1}{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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{49}{4}$$

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

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{49}{4}}$$

Whakarūnātia.

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

Me tango \frac{1}{2} mai i ngā taha e rua o te whārite.

$$x=3 x=-4$$