Whakaoti i te x^2-x-6=0

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

Whakaoti mō x

$$x=-2$$

$$x=3$$

Ngā Upane Whakamahi Tauwehe

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

$$a+b=-1 ab=-6$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -6.

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

Tātaihia te tapeke mō ia takirua.

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

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

$$a=-3 b=2$$

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

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.

$$a+b=-1 ab=1\left(-6\right)=-6$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -6.

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

Tātaihia te tapeke mō ia takirua.

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

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

$$a=-3 b=2$$

Tuhia anō te x^{2}-x-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.

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

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

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-6\right)}}{2}$$

Whakareatia -4 ki te -6.

$$x=\frac{-\left(-1\right)±\sqrt{1+24}}{2}$$

Tāpiri 1 ki te 24.

$$x=\frac{-\left(-1\right)±\sqrt{25}}{2}$$

Tuhia te pūtakerua o te 25.

$$x=\frac{-\left(-1\right)±5}{2}$$

Ko te tauaro o -1 ko 1.

$$x=\frac{1±5}{2}$$

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

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

Whakawehe 6 ki te 2.

$$x=3$$

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

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

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.

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

Me tāpiri 6 ki ngā taha e rua o te whārite.

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

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

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

Tango -6 mai i 0.

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

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.

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

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

$$x^{2}-x+\frac{1}{4}=6+\frac{1}{4}$$

Tāpiri 6 ki te \frac{1}{4}.

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

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{25}{4}$$

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

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Whakarūnātia.

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

Me tāpiri \frac{1}{2} ki ngā taha e rua o te whārite.

$$x=3 x=-2$$