Whakaoti i te 10y^2+3y-4

$$10 y ^ { 2 } + 3 y - 4$$

Tauwehe

$$\left(2y-1\right)\left(5y+4\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 10y^{2}+ay+by-4. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=3 ab=10\left(-4\right)=-40$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōrunga te a+b, he nui ake te uara pū o te tau tōrunga i tō te tōraro. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -40.

$$-1,40 -2,20 -4,10 -5,8$$

Tātaihia te tapeke mō ia takirua.

$$-1+40=39 -2+20=18 -4+10=6 -5+8=3$$

Ko te otinga te takirua ka hoatu i te tapeke 3.

$$a=-5 b=8$$

Tuhia anō te 10y^{2}+3y-4 hei \left(10y^{2}-5y\right)+\left(8y-4\right).

$$\left(10y^{2}-5y\right)+\left(8y-4\right)$$

Tauwehea te 5y i te tuatahi me te 4 i te rōpū tuarua.

$$5y\left(2y-1\right)+4\left(2y-1\right)$$

Whakatauwehea atu te kīanga pātahi 2y-1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2y-1\right)\left(5y+4\right)$$

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.

$$10y^{2}+3y-4=0$$

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.

$$y=\frac{-3±\sqrt{3^{2}-4\times 10\left(-4\right)}}{2\times 10}$$

Pūrua 3.

$$y=\frac{-3±\sqrt{9-4\times 10\left(-4\right)}}{2\times 10}$$

Whakareatia -4 ki te 10.

$$y=\frac{-3±\sqrt{9-40\left(-4\right)}}{2\times 10}$$

Whakareatia -40 ki te -4.

$$y=\frac{-3±\sqrt{9+160}}{2\times 10}$$

Tāpiri 9 ki te 160.

$$y=\frac{-3±\sqrt{169}}{2\times 10}$$

Tuhia te pūtakerua o te 169.

$$y=\frac{-3±13}{2\times 10}$$

Whakareatia 2 ki te 10.

$$y=\frac{-3±13}{20}$$

Nā, me whakaoti te whārite y=\frac{-3±13}{20} ina he tāpiri te ±. Tāpiri -3 ki te 13.

$$y=\frac{10}{20}$$

Whakahekea te hautanga \frac{10}{20} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 10.

$$y=\frac{1}{2}$$

Nā, me whakaoti te whārite y=\frac{-3±13}{20} ina he tango te ±. Tango 13 mai i -3.

$$y=-\frac{16}{20}$$

Whakahekea te hautanga \frac{-16}{20} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$y=-\frac{4}{5}$$

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

$$10y^{2}+3y-4=10\left(y-\frac{1}{2}\right)\left(y-\left(-\frac{4}{5}\right)\right)$$

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

$$10y^{2}+3y-4=10\left(y-\frac{1}{2}\right)\left(y+\frac{4}{5}\right)$$

Tango \frac{1}{2} mai i y 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.

$$10y^{2}+3y-4=10\times \frac{2y-1}{2}\left(y+\frac{4}{5}\right)$$

Tāpiri \frac{4}{5} ki te y 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.

$$10y^{2}+3y-4=10\times \frac{2y-1}{2}\times \frac{5y+4}{5}$$

Whakareatia \frac{2y-1}{2} ki te \frac{5y+4}{5} mā te whakarea taurunga ki te taurunga me te tauraro ki te tauraro, ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$10y^{2}+3y-4=10\times \frac{\left(2y-1\right)\left(5y+4\right)}{2\times 5}$$

Whakareatia 2 ki te 5.

$$10y^{2}+3y-4=10\times \frac{\left(2y-1\right)\left(5y+4\right)}{10}$$

Whakakorea atu te tauwehe pūnoa nui rawa 10 i roto i te 10 me te 10.

$$10y^{2}+3y-4=\left(2y-1\right)\left(5y+4\right)$$

Aromātai

$$\left(2y-1\right)\left(5y+4\right)$$