Whakaoti i te 4a^2-19a-5

$$4 { a }^{ 2 } -19a-5$$

Tauwehe

$$\left(a-5\right)\left(4a+1\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 4a^{2}+pa+qa-5. Hei kimi p me q, whakaritea tētahi pūnaha kia whakaoti.

$$p+q=-19 pq=4\left(-5\right)=-20$$

I te mea kua tōraro te pq, he tauaro ngā tohu o p me q. I te mea kua tōraro te p+q, he nui ake te uara pū o te tau tōraro i tō te tōrunga. 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.

$$1-20=-19 2-10=-8 4-5=-1$$

Ko te otinga te takirua ka hoatu i te tapeke -19.

$$p=-20 q=1$$

Tuhia anō te 4a^{2}-19a-5 hei \left(4a^{2}-20a\right)+\left(a-5\right).

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

Whakatauwehea atu 4a i te 4a^{2}-20a.

$$4a\left(a-5\right)+a-5$$

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

$$\left(a-5\right)\left(4a+1\right)$$

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.

$$4a^{2}-19a-5=0$$

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.

$$a=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 4\left(-5\right)}}{2\times 4}$$

Pūrua -19.

$$a=\frac{-\left(-19\right)±\sqrt{361-4\times 4\left(-5\right)}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$a=\frac{-\left(-19\right)±\sqrt{361-16\left(-5\right)}}{2\times 4}$$

Whakareatia -16 ki te -5.

$$a=\frac{-\left(-19\right)±\sqrt{361+80}}{2\times 4}$$

Tāpiri 361 ki te 80.

$$a=\frac{-\left(-19\right)±\sqrt{441}}{2\times 4}$$

Tuhia te pūtakerua o te 441.

$$a=\frac{-\left(-19\right)±21}{2\times 4}$$

Ko te tauaro o -19 ko 19.

$$a=\frac{19±21}{2\times 4}$$

Whakareatia 2 ki te 4.

$$a=\frac{19±21}{8}$$

Nā, me whakaoti te whārite a=\frac{19±21}{8} ina he tāpiri te ±. Tāpiri 19 ki te 21.

$$a=\frac{40}{8}$$

Whakawehe 40 ki te 8.

$$a=5$$

Nā, me whakaoti te whārite a=\frac{19±21}{8} ina he tango te ±. Tango 21 mai i 19.

$$a=-\frac{2}{8}$$

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

$$a=-\frac{1}{4}$$

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 -\frac{1}{4} mō te x_{2}.

$$4a^{2}-19a-5=4\left(a-5\right)\left(a-\left(-\frac{1}{4}\right)\right)$$

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

$$4a^{2}-19a-5=4\left(a-5\right)\left(a+\frac{1}{4}\right)$$

Tāpiri \frac{1}{4} ki te a mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$4a^{2}-19a-5=4\left(a-5\right)\times \frac{4a+1}{4}$$

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

$$4a^{2}-19a-5=\left(a-5\right)\left(4a+1\right)$$

Aromātai

$$\left(a-5\right)\left(4a+1\right)$$