Whakaoti i te 4c^2-9c+5

$$4 c ^ { 2 } - 9 c + 5$$

Tauwehe

$$\left(c-1\right)\left(4c-5\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 4c^{2}+ac+bc+5. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-9 ab=4\times 5=20$$

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 4c^{2}-9c+5 hei \left(4c^{2}-5c\right)+\left(-4c+5\right).

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

Tauwehea te c i te tuatahi me te -1 i te rōpū tuarua.

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

Whakatauwehea atu te kīanga pātahi 4c-5 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(4c-5\right)\left(c-1\right)$$

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.

$$4c^{2}-9c+5=0$$

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.

$$c=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\times 4\times 5}}{2\times 4}$$

Pūrua -9.

$$c=\frac{-\left(-9\right)±\sqrt{81-4\times 4\times 5}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$c=\frac{-\left(-9\right)±\sqrt{81-16\times 5}}{2\times 4}$$

Whakareatia -16 ki te 5.

$$c=\frac{-\left(-9\right)±\sqrt{81-80}}{2\times 4}$$

Tāpiri 81 ki te -80.

$$c=\frac{-\left(-9\right)±\sqrt{1}}{2\times 4}$$

Tuhia te pūtakerua o te 1.

$$c=\frac{-\left(-9\right)±1}{2\times 4}$$

Ko te tauaro o -9 ko 9.

$$c=\frac{9±1}{2\times 4}$$

Whakareatia 2 ki te 4.

$$c=\frac{9±1}{8}$$

Nā, me whakaoti te whārite c=\frac{9±1}{8} ina he tāpiri te ±. Tāpiri 9 ki te 1.

$$c=\frac{10}{8}$$

Whakahekea te hautanga \frac{10}{8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$c=\frac{5}{4}$$

Nā, me whakaoti te whārite c=\frac{9±1}{8} ina he tango te ±. Tango 1 mai i 9.

$$c=\frac{8}{8}$$

Whakawehe 8 ki te 8.

$$c=1$$

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 \frac{5}{4} mō te x_{1} me te 1 mō te x_{2}.

$$4c^{2}-9c+5=4\left(c-\frac{5}{4}\right)\left(c-1\right)$$

Tango \frac{5}{4} mai i c mā te kimi i te tauraro pātahi me te tango i ngā taurunga, ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$4c^{2}-9c+5=4\times \frac{4c-5}{4}\left(c-1\right)$$

Whakakorea atu te tauwehe pūnoa nui rawa 4 i roto i te 4 me te 4.

$$4c^{2}-9c+5=\left(4c-5\right)\left(c-1\right)$$

Aromātai

$$\left(c-1\right)\left(4c-5\right)$$