Whakaoti i te x^2-1=x+1

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

Whakaoti mō x

$$x=-1$$

$$x=2$$

Ngā Upane Whakamahi Tauwehe

Tangohia te x mai i ngā taha e rua.

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

Tangohia te 1 mai i ngā taha e rua.

$$x^{2}-1-x-1=0$$

Tangohia te 1 i te -1, ka -2.

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

Hurinahatia te pūrau ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

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

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

$$a+b=-1 ab=-2$$

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. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=-2 b=1$$

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

$$\left(x-2\right)\left(x+1\right)$$

Hei kimi otinga whārite, me whakaoti te x-2=0 me te x+1=0.

$$x=2 x=-1$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Tangohia te x mai i ngā taha e rua.

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

Tangohia te 1 mai i ngā taha e rua.

$$x^{2}-1-x-1=0$$

Tangohia te 1 i te -1, ka -2.

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

Hurinahatia te pūrau ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

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

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

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

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. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=-2 b=1$$

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

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

Whakatauwehea atu x i te x^{2}-2x.

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

Whakatauwehea atu te kīanga pātahi x-2 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-2\right)\left(x+1\right)$$

Hei kimi otinga whārite, me whakaoti te x-2=0 me te x+1=0.

$$x=2 x=-1$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te x mai i ngā taha e rua.

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

Tangohia te 1 mai i ngā taha e rua.

$$x^{2}-1-x-1=0$$

Tangohia te 1 i te -1, ka -2.

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

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.

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

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

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

Whakareatia -4 ki te -2.

$$x=\frac{-\left(-1\right)±\sqrt{1+8}}{2}$$

Tāpiri 1 ki te 8.

$$x=\frac{-\left(-1\right)±\sqrt{9}}{2}$$

Tuhia te pūtakerua o te 9.

$$x=\frac{-\left(-1\right)±3}{2}$$

Ko te tauaro o -1 ko 1.

$$x=\frac{1±3}{2}$$

Nā, me whakaoti te whārite x=\frac{1±3}{2} ina he tāpiri te ±. Tāpiri 1 ki te 3.

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

Whakawehe 4 ki te 2.

$$x=2$$

Nā, me whakaoti te whārite x=\frac{1±3}{2} ina he tango te ±. Tango 3 mai i 1.

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

Whakawehe -2 ki te 2.

$$x=-1$$

Kua oti te whārite te whakatau.

$$x=2 x=-1$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te x mai i ngā taha e rua.

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

Me tāpiri te 1 ki ngā taha e rua.

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

Tāpirihia te 1 ki te 1, ka 2.

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

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.

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

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

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

Tāpiri 2 ki te \frac{1}{4}.

$$x^{2}-x+\frac{1}{4}=\frac{9}{4}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{9}{4}$$

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

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Whakarūnātia.

$$x-\frac{1}{2}=\frac{3}{2} x-\frac{1}{2}=-\frac{3}{2}$$

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

$$x=2 x=-1$$