Whakaoti i te x^2+9x+8

$$x ^ { 2 } + 9 x + 8$$

Tauwehe

$$\left(x+1\right)\left(x+8\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei x^{2}+ax+bx+8. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=9 ab=1\times 8=8$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōrunga te a+b, he tōrunga hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 8.

$$1,8 2,4$$

Tātaihia te tapeke mō ia takirua.

$$1+8=9 2+4=6$$

Ko te otinga te takirua ka hoatu i te tapeke 9.

$$a=1 b=8$$

Tuhia anō te x^{2}+9x+8 hei \left(x^{2}+x\right)+\left(8x+8\right).

$$\left(x^{2}+x\right)+\left(8x+8\right)$$

Tauwehea te x i te tuatahi me te 8 i te rōpū tuarua.

$$x\left(x+1\right)+8\left(x+1\right)$$

Whakatauwehea atu te kīanga pātahi x+1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x+1\right)\left(x+8\right)$$

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.

$$x^{2}+9x+8=0$$

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.

$$x=\frac{-9±\sqrt{9^{2}-4\times 8}}{2}$$

Pūrua 9.

$$x=\frac{-9±\sqrt{81-4\times 8}}{2}$$

Whakareatia -4 ki te 8.

$$x=\frac{-9±\sqrt{81-32}}{2}$$

Tāpiri 81 ki te -32.

$$x=\frac{-9±\sqrt{49}}{2}$$

Tuhia te pūtakerua o te 49.

$$x=\frac{-9±7}{2}$$

Nā, me whakaoti te whārite x=\frac{-9±7}{2} ina he tāpiri te ±. Tāpiri -9 ki te 7.

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

Whakawehe -2 ki te 2.

$$x=-1$$

Nā, me whakaoti te whārite x=\frac{-9±7}{2} ina he tango te ±. Tango 7 mai i -9.

$$x=-\frac{16}{2}$$

Whakawehe -16 ki te 2.

$$x=-8$$

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 -1 mō te x_{1} me te -8 mō te x_{2}.

$$x^{2}+9x+8=\left(x-\left(-1\right)\right)\left(x-\left(-8\right)\right)$$

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

$$x^{2}+9x+8=\left(x+1\right)\left(x+8\right)$$

Aromātai

$$\left(x+1\right)\left(x+8\right)$$