Whakaoti i te 5y^2+27y+10

$$5 y ^ { 2 } + 27 y + 10$$

Tauwehe

$$\left(y+5\right)\left(5y+2\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 5y^{2}+ay+by+10. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=27 ab=5\times 10=50$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōrunga te a+b, he tōrunga hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 50.

$$1,50 2,25 5,10$$

Tātaihia te tapeke mō ia takirua.

$$1+50=51 2+25=27 5+10=15$$

Ko te otinga te takirua ka hoatu i te tapeke 27.

$$a=2 b=25$$

Tuhia anō te 5y^{2}+27y+10 hei \left(5y^{2}+2y\right)+\left(25y+10\right).

$$\left(5y^{2}+2y\right)+\left(25y+10\right)$$

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

$$y\left(5y+2\right)+5\left(5y+2\right)$$

Whakatauwehea atu te kīanga pātahi 5y+2 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(5y+2\right)\left(y+5\right)$$

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.

$$5y^{2}+27y+10=0$$

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.

$$y=\frac{-27±\sqrt{27^{2}-4\times 5\times 10}}{2\times 5}$$

Pūrua 27.

$$y=\frac{-27±\sqrt{729-4\times 5\times 10}}{2\times 5}$$

Whakareatia -4 ki te 5.

$$y=\frac{-27±\sqrt{729-20\times 10}}{2\times 5}$$

Whakareatia -20 ki te 10.

$$y=\frac{-27±\sqrt{729-200}}{2\times 5}$$

Tāpiri 729 ki te -200.

$$y=\frac{-27±\sqrt{529}}{2\times 5}$$

Tuhia te pūtakerua o te 529.

$$y=\frac{-27±23}{2\times 5}$$

Whakareatia 2 ki te 5.

$$y=\frac{-27±23}{10}$$

Nā, me whakaoti te whārite y=\frac{-27±23}{10} ina he tāpiri te ±. Tāpiri -27 ki te 23.

$$y=-\frac{4}{10}$$

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

$$y=-\frac{2}{5}$$

Nā, me whakaoti te whārite y=\frac{-27±23}{10} ina he tango te ±. Tango 23 mai i -27.

$$y=-\frac{50}{10}$$

Whakawehe -50 ki te 10.

$$y=-5$$

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{2}{5} mō te x_{1} me te -5 mō te x_{2}.

$$5y^{2}+27y+10=5\left(y-\left(-\frac{2}{5}\right)\right)\left(y-\left(-5\right)\right)$$

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

$$5y^{2}+27y+10=5\left(y+\frac{2}{5}\right)\left(y+5\right)$$

Tāpiri \frac{2}{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.

$$5y^{2}+27y+10=5\times \frac{5y+2}{5}\left(y+5\right)$$

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

$$5y^{2}+27y+10=\left(5y+2\right)\left(y+5\right)$$

Aromātai

$$\left(y+5\right)\left(5y+2\right)$$