Whakaoti i te 2q^2-7q+5

$$2 q ^ { 2 } - 7 q + 5$$

Tauwehe

$$\left(q-1\right)\left(2q-5\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 2q^{2}+aq+bq+5. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-7 ab=2\times 5=10$$

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 10.

$$-1,-10 -2,-5$$

Tātaihia te tapeke mō ia takirua.

$$-1-10=-11 -2-5=-7$$

Ko te otinga te takirua ka hoatu i te tapeke -7.

$$a=-5 b=-2$$

Tuhia anō te 2q^{2}-7q+5 hei \left(2q^{2}-5q\right)+\left(-2q+5\right).

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

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

$$q\left(2q-5\right)-\left(2q-5\right)$$

Whakatauwehea atu te kīanga pātahi 2q-5 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2q-5\right)\left(q-1\right)$$

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.

$$2q^{2}-7q+5=0$$

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.

$$q=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 2\times 5}}{2\times 2}$$

Pūrua -7.

$$q=\frac{-\left(-7\right)±\sqrt{49-4\times 2\times 5}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$q=\frac{-\left(-7\right)±\sqrt{49-8\times 5}}{2\times 2}$$

Whakareatia -8 ki te 5.

$$q=\frac{-\left(-7\right)±\sqrt{49-40}}{2\times 2}$$

Tāpiri 49 ki te -40.

$$q=\frac{-\left(-7\right)±\sqrt{9}}{2\times 2}$$

Tuhia te pūtakerua o te 9.

$$q=\frac{-\left(-7\right)±3}{2\times 2}$$

Ko te tauaro o -7 ko 7.

$$q=\frac{7±3}{2\times 2}$$

Whakareatia 2 ki te 2.

$$q=\frac{7±3}{4}$$

Nā, me whakaoti te whārite q=\frac{7±3}{4} ina he tāpiri te ±. Tāpiri 7 ki te 3.

$$q=\frac{10}{4}$$

Whakahekea te hautanga \frac{10}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$q=\frac{5}{2}$$

Nā, me whakaoti te whārite q=\frac{7±3}{4} ina he tango te ±. Tango 3 mai i 7.

$$q=\frac{4}{4}$$

Whakawehe 4 ki te 4.

$$q=1$$

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}{2} mō te x_{1} me te 1 mō te x_{2}.

$$2q^{2}-7q+5=2\left(q-\frac{5}{2}\right)\left(q-1\right)$$

Tango \frac{5}{2} mai i q 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.

$$2q^{2}-7q+5=2\times \frac{2q-5}{2}\left(q-1\right)$$

Whakakorea atu te tauwehe pūnoa nui rawa 2 i roto i te 2 me te 2.

$$2q^{2}-7q+5=\left(2q-5\right)\left(q-1\right)$$

Aromātai

$$\left(q-1\right)\left(2q-5\right)$$