Whakaoti i te 5a^2-16a+3

$$5 a ^ { 2 } - 16 a + 3$$

Tauwehe

$$\left(a-3\right)\left(5a-1\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 5a^{2}+pa+qa+3. Hei kimi p me q, whakaritea tētahi pūnaha kia whakaoti.

$$p+q=-16 pq=5\times 3=15$$

I te mea kua tōrunga te pq, he ōrite te tohu o p me q. I te mea kua tōraro te p+q, he tōraro hoki a p me q. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 15.

$$-1,-15 -3,-5$$

Tātaihia te tapeke mō ia takirua.

$$-1-15=-16 -3-5=-8$$

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

$$p=-15 q=-1$$

Tuhia anō te 5a^{2}-16a+3 hei \left(5a^{2}-15a\right)+\left(-a+3\right).

$$\left(5a^{2}-15a\right)+\left(-a+3\right)$$

Tauwehea te 5a i te tuatahi me te -1 i te rōpū tuarua.

$$5a\left(a-3\right)-\left(a-3\right)$$

Whakatauwehea atu te kīanga pātahi a-3 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(a-3\right)\left(5a-1\right)$$

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.

$$5a^{2}-16a+3=0$$

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.

$$a=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 5\times 3}}{2\times 5}$$

Pūrua -16.

$$a=\frac{-\left(-16\right)±\sqrt{256-4\times 5\times 3}}{2\times 5}$$

Whakareatia -4 ki te 5.

$$a=\frac{-\left(-16\right)±\sqrt{256-20\times 3}}{2\times 5}$$

Whakareatia -20 ki te 3.

$$a=\frac{-\left(-16\right)±\sqrt{256-60}}{2\times 5}$$

Tāpiri 256 ki te -60.

$$a=\frac{-\left(-16\right)±\sqrt{196}}{2\times 5}$$

Tuhia te pūtakerua o te 196.

$$a=\frac{-\left(-16\right)±14}{2\times 5}$$

Ko te tauaro o -16 ko 16.

$$a=\frac{16±14}{2\times 5}$$

Whakareatia 2 ki te 5.

$$a=\frac{16±14}{10}$$

Nā, me whakaoti te whārite a=\frac{16±14}{10} ina he tāpiri te ±. Tāpiri 16 ki te 14.

$$a=\frac{30}{10}$$

Whakawehe 30 ki te 10.

$$a=3$$

Nā, me whakaoti te whārite a=\frac{16±14}{10} ina he tango te ±. Tango 14 mai i 16.

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

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

$$a=\frac{1}{5}$$

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

$$5a^{2}-16a+3=5\left(a-3\right)\left(a-\frac{1}{5}\right)$$

Tango \frac{1}{5} mai i a 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.

$$5a^{2}-16a+3=5\left(a-3\right)\times \frac{5a-1}{5}$$

Whakakorea atu te tauwehe pūnoa nui rawa 5 i roto i te 5 me te 5.

$$5a^{2}-16a+3=\left(a-3\right)\left(5a-1\right)$$

Aromātai

$$\left(a-3\right)\left(5a-1\right)$$