Whakaoti i te s^2-9s+20

$$s ^ { 2 } - 9 s + 20$$

Tauwehe

$$\left(s-5\right)\left(s-4\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei s^{2}+as+bs+20. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-9 ab=1\times 20=20$$

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 20.

$$-1,-20 -2,-10 -4,-5$$

Tātaihia te tapeke mō ia takirua.

$$-1-20=-21 -2-10=-12 -4-5=-9$$

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

$$a=-5 b=-4$$

Tuhia anō te s^{2}-9s+20 hei \left(s^{2}-5s\right)+\left(-4s+20\right).

$$\left(s^{2}-5s\right)+\left(-4s+20\right)$$

Tauwehea te s i te tuatahi me te -4 i te rōpū tuarua.

$$s\left(s-5\right)-4\left(s-5\right)$$

Whakatauwehea atu te kīanga pātahi s-5 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(s-5\right)\left(s-4\right)$$

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.

$$s^{2}-9s+20=0$$

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.

$$s=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 20}}{2}$$

Pūrua -9.

$$s=\frac{-\left(-9\right)±\sqrt{81-4\times 20}}{2}$$

Whakareatia -4 ki te 20.

$$s=\frac{-\left(-9\right)±\sqrt{81-80}}{2}$$

Tāpiri 81 ki te -80.

$$s=\frac{-\left(-9\right)±\sqrt{1}}{2}$$

Tuhia te pūtakerua o te 1.

$$s=\frac{-\left(-9\right)±1}{2}$$

Ko te tauaro o -9 ko 9.

$$s=\frac{9±1}{2}$$

Nā, me whakaoti te whārite s=\frac{9±1}{2} ina he tāpiri te ±. Tāpiri 9 ki te 1.

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

Whakawehe 10 ki te 2.

$$s=5$$

Nā, me whakaoti te whārite s=\frac{9±1}{2} ina he tango te ±. Tango 1 mai i 9.

$$s=\frac{8}{2}$$

Whakawehe 8 ki te 2.

$$s=4$$

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 5 mō te x_{1} me te 4 mō te x_{2}.

$$s^{2}-9s+20=\left(s-5\right)\left(s-4\right)$$

Aromātai

$$\left(s-5\right)\left(s-4\right)$$