Whakaoti i te y^2+8y+12=

$$y ^ { 2 } + 8 y + 12 =$$

Tauwehe

$$\left(y+2\right)\left(y+6\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei y^{2}+ay+by+12. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=8 ab=1\times 12=12$$

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 12.

$$1,12 2,6 3,4$$

Tātaihia te tapeke mō ia takirua.

$$1+12=13 2+6=8 3+4=7$$

Ko te otinga te takirua ka hoatu i te tapeke 8.

$$a=2 b=6$$

Tuhia anō te y^{2}+8y+12 hei \left(y^{2}+2y\right)+\left(6y+12\right).

$$\left(y^{2}+2y\right)+\left(6y+12\right)$$

Tauwehea te y i te tuatahi me te 6 i te rōpū tuarua.

$$y\left(y+2\right)+6\left(y+2\right)$$

Whakatauwehea atu te kīanga pātahi y+2 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(y+2\right)\left(y+6\right)$$

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.

$$y^{2}+8y+12=0$$

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.

$$y=\frac{-8±\sqrt{8^{2}-4\times 12}}{2}$$

Pūrua 8.

$$y=\frac{-8±\sqrt{64-4\times 12}}{2}$$

Whakareatia -4 ki te 12.

$$y=\frac{-8±\sqrt{64-48}}{2}$$

Tāpiri 64 ki te -48.

$$y=\frac{-8±\sqrt{16}}{2}$$

Tuhia te pūtakerua o te 16.

$$y=\frac{-8±4}{2}$$

Nā, me whakaoti te whārite y=\frac{-8±4}{2} ina he tāpiri te ±. Tāpiri -8 ki te 4.

$$y=-\frac{4}{2}$$

Whakawehe -4 ki te 2.

$$y=-2$$

Nā, me whakaoti te whārite y=\frac{-8±4}{2} ina he tango te ±. Tango 4 mai i -8.

$$y=-\frac{12}{2}$$

Whakawehe -12 ki te 2.

$$y=-6$$

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 -2 mō te x_{1} me te -6 mō te x_{2}.

$$y^{2}+8y+12=\left(y-\left(-2\right)\right)\left(y-\left(-6\right)\right)$$

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

$$y^{2}+8y+12=\left(y+2\right)\left(y+6\right)$$

Aromātai

$$\left(y+2\right)\left(y+6\right)$$