Whakaoti i te -30+x^2-x

$$-30+ { x }^{ 2 } -x$$

Tauwehe

$$\left(x-6\right)\left(x+5\right)$$

Ngā Upane Otinga

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

$$x^{2}-x-30$$

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei x^{2}+ax+bx-30. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-1 ab=1\left(-30\right)=-30$$

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 -30.

$$1,-30 2,-15 3,-10 5,-6$$

Tātaihia te tapeke mō ia takirua.

$$1-30=-29 2-15=-13 3-10=-7 5-6=-1$$

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

$$a=-6 b=5$$

Tuhia anō te x^{2}-x-30 hei \left(x^{2}-6x\right)+\left(5x-30\right).

$$\left(x^{2}-6x\right)+\left(5x-30\right)$$

Tauwehea te x i te tuatahi me te 5 i te rōpū tuarua.

$$x\left(x-6\right)+5\left(x-6\right)$$

Whakatauwehea atu te kīanga pātahi x-6 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-6\right)\left(x+5\right)$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Ka taea te huamaha pūrua te tauwehe mā te whakamahi i te huringa ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), ina ko x_{1} me x_{2} ngā otinga o te whārite pūrua ax^{2}+bx+c=0.

$$x^{2}-x-30=0$$

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.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-30\right)}}{2}$$

Whakareatia -4 ki te -30.

$$x=\frac{-\left(-1\right)±\sqrt{1+120}}{2}$$

Tāpiri 1 ki te 120.

$$x=\frac{-\left(-1\right)±\sqrt{121}}{2}$$

Tuhia te pūtakerua o te 121.

$$x=\frac{-\left(-1\right)±11}{2}$$

Ko te tauaro o -1 ko 1.

$$x=\frac{1±11}{2}$$

Nā, me whakaoti te whārite x=\frac{1±11}{2} ina he tāpiri te ±. Tāpiri 1 ki te 11.

$$x=\frac{12}{2}$$

Whakawehe 12 ki te 2.

$$x=6$$

Nā, me whakaoti te whārite x=\frac{1±11}{2} ina he tango te ±. Tango 11 mai i 1.

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

Whakawehe -10 ki te 2.

$$x=-5$$

Tauwehea te kīanga taketake mā te whakamahi i te ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Me whakakapi te 6 mō te x_{1} me te -5 mō te x_{2}.

$$x^{2}-x-30=\left(x-6\right)\left(x-\left(-5\right)\right)$$

Whakamāmātia ngā kīanga katoa o te āhua p-\left(-q\right) ki te p+q.

$$x^{2}-x-30=\left(x-6\right)\left(x+5\right)$$

Aromātai

$$\left(x-6\right)\left(x+5\right)$$