Whakaoti i te 4=x^2-2x-11

$$4 = x ^ { 2 } - 2 x - 11$$

Whakaoti mō x

$$x=-3$$

$$x=5$$

Ngā Upane Whakamahi Tauwehe

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$x^{2}-2x-11=4$$

Tangohia te 4 mai i ngā taha e rua.

$$x^{2}-2x-11-4=0$$

Tangohia te 4 i te -11, ka -15.

$$x^{2}-2x-15=0$$

Hei whakaoti i te whārite, whakatauwehea te x^{2}-2x-15 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.

$$a+b=-2 ab=-15$$

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 -15.

$$1,-15 3,-5$$

Tātaihia te tapeke mō ia takirua.

$$1-15=-14 3-5=-2$$

Ko te otinga te takirua ka hoatu i te tapeke -2.

$$a=-5 b=3$$

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

$$\left(x-5\right)\left(x+3\right)$$

Hei kimi otinga whārite, me whakaoti te x-5=0 me te x+3=0.

$$x=5 x=-3$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$x^{2}-2x-11=4$$

Tangohia te 4 mai i ngā taha e rua.

$$x^{2}-2x-11-4=0$$

Tangohia te 4 i te -11, ka -15.

$$x^{2}-2x-15=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei x^{2}+ax+bx-15. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-2 ab=1\left(-15\right)=-15$$

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 -15.

$$1,-15 3,-5$$

Tātaihia te tapeke mō ia takirua.

$$1-15=-14 3-5=-2$$

Ko te otinga te takirua ka hoatu i te tapeke -2.

$$a=-5 b=3$$

Tuhia anō te x^{2}-2x-15 hei \left(x^{2}-5x\right)+\left(3x-15\right).

$$\left(x^{2}-5x\right)+\left(3x-15\right)$$

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

$$x\left(x-5\right)+3\left(x-5\right)$$

Whakatauwehea atu te kīanga pātahi x-5 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-5\right)\left(x+3\right)$$

Hei kimi otinga whārite, me whakaoti te x-5=0 me te x+3=0.

$$x=5 x=-3$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$x^{2}-2x-11=4$$

Tangohia te 4 mai i ngā taha e rua.

$$x^{2}-2x-11-4=0$$

Tangohia te 4 i te -11, ka -15.

$$x^{2}-2x-15=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -2 mō b, me -15 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-15\right)}}{2}$$

Pūrua -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-15\right)}}{2}$$

Whakareatia -4 ki te -15.

$$x=\frac{-\left(-2\right)±\sqrt{4+60}}{2}$$

Tāpiri 4 ki te 60.

$$x=\frac{-\left(-2\right)±\sqrt{64}}{2}$$

Tuhia te pūtakerua o te 64.

$$x=\frac{-\left(-2\right)±8}{2}$$

Ko te tauaro o -2 ko 2.

$$x=\frac{2±8}{2}$$

Nā, me whakaoti te whārite x=\frac{2±8}{2} ina he tāpiri te ±. Tāpiri 2 ki te 8.

$$x=\frac{10}{2}$$

Whakawehe 10 ki te 2.

$$x=5$$

Nā, me whakaoti te whārite x=\frac{2±8}{2} ina he tango te ±. Tango 8 mai i 2.

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

Whakawehe -6 ki te 2.

$$x=-3$$

Kua oti te whārite te whakatau.

$$x=5 x=-3$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$x^{2}-2x-11=4$$

Me tāpiri te 11 ki ngā taha e rua.

$$x^{2}-2x=4+11$$

Tāpirihia te 4 ki te 11, ka 15.

$$x^{2}-2x=15$$

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

$$x^{2}-2x+1=15+1$$

Tāpiri 15 ki te 1.

$$x^{2}-2x+1=16$$

Tauwehea x^{2}-2x+1. 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}.

$$\left(x-1\right)^{2}=16$$

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

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{16}$$

Whakarūnātia.

$$x-1=4 x-1=-4$$

Me tāpiri 1 ki ngā taha e rua o te whārite.

$$x=5 x=-3$$