Whakaoti i te y^2-16y+60

$$y ^ { 2 } - 16 y + 60$$

Tauwehe

$$\left(y-10\right)\left(y-6\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei y^{2}+ay+by+60. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-16 ab=1\times 60=60$$

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 60.

$$-1,-60 -2,-30 -3,-20 -4,-15 -5,-12 -6,-10$$

Tātaihia te tapeke mō ia takirua.

$$-1-60=-61 -2-30=-32 -3-20=-23 -4-15=-19 -5-12=-17 -6-10=-16$$

Ko te otinga te takirua ka hoatu i te tapeke -16.

$$a=-10 b=-6$$

Tuhia anō te y^{2}-16y+60 hei \left(y^{2}-10y\right)+\left(-6y+60\right).

$$\left(y^{2}-10y\right)+\left(-6y+60\right)$$

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

$$y\left(y-10\right)-6\left(y-10\right)$$

Whakatauwehea atu te kīanga pātahi y-10 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(y-10\right)\left(y-6\right)$$

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.

$$y^{2}-16y+60=0$$

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.

$$y=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 60}}{2}$$

Pūrua -16.

$$y=\frac{-\left(-16\right)±\sqrt{256-4\times 60}}{2}$$

Whakareatia -4 ki te 60.

$$y=\frac{-\left(-16\right)±\sqrt{256-240}}{2}$$

Tāpiri 256 ki te -240.

$$y=\frac{-\left(-16\right)±\sqrt{16}}{2}$$

Tuhia te pūtakerua o te 16.

$$y=\frac{-\left(-16\right)±4}{2}$$

Ko te tauaro o -16 ko 16.

$$y=\frac{16±4}{2}$$

Nā, me whakaoti te whārite y=\frac{16±4}{2} ina he tāpiri te ±. Tāpiri 16 ki te 4.

$$y=\frac{20}{2}$$

Whakawehe 20 ki te 2.

$$y=10$$

Nā, me whakaoti te whārite y=\frac{16±4}{2} ina he tango te ±. Tango 4 mai i 16.

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

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 10 mō te x_{1} me te 6 mō te x_{2}.

$$y^{2}-16y+60=\left(y-10\right)\left(y-6\right)$$

Aromātai

$$\left(y-10\right)\left(y-6\right)$$