Whakaoti i te x^2-11x+30

$${ x }^{ 2 } -11x+30$$

Tauwehe

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

Ngā Upane Otinga

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.

$$a+b=-11 ab=1\times 30=30$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōraro te a+b, he tōraro hoki a a me b. 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.

$$-1-30=-31 -2-15=-17 -3-10=-13 -5-6=-11$$

Ko te otinga te takirua ka hoatu i te tapeke -11.

$$a=-6 b=-5$$

Tuhia anō te x^{2}-11x+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.

$$x^{2}-11x+30=0$$

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.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 30}}{2}$$

Pūrua -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 30}}{2}$$

Whakareatia -4 ki te 30.

$$x=\frac{-\left(-11\right)±\sqrt{121-120}}{2}$$

Tāpiri 121 ki te -120.

$$x=\frac{-\left(-11\right)±\sqrt{1}}{2}$$

Tuhia te pūtakerua o te 1.

$$x=\frac{-\left(-11\right)±1}{2}$$

Ko te tauaro o -11 ko 11.

$$x=\frac{11±1}{2}$$

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

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

Whakawehe 12 ki te 2.

$$x=6$$

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

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

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}.

$$x^{2}-11x+30=\left(x-6\right)\left(x-5\right)$$

Aromātai

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