Whakaoti i te 6y^2+5y-4

$$6 y ^ { 2 } + 5 y - 4$$

Tauwehe

$$\left(2y-1\right)\left(3y+4\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 6y^{2}+ay+by-4. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=5 ab=6\left(-4\right)=-24$$

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 -24.

$$-1,24 -2,12 -3,8 -4,6$$

Tātaihia te tapeke mō ia takirua.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

Ko te otinga te takirua ka hoatu i te tapeke 5.

$$a=-3 b=8$$

Tuhia anō te 6y^{2}+5y-4 hei \left(6y^{2}-3y\right)+\left(8y-4\right).

$$\left(6y^{2}-3y\right)+\left(8y-4\right)$$

Tauwehea te 3y i te tuatahi me te 4 i te rōpū tuarua.

$$3y\left(2y-1\right)+4\left(2y-1\right)$$

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

$$\left(2y-1\right)\left(3y+4\right)$$

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.

$$6y^{2}+5y-4=0$$

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.

$$y=\frac{-5±\sqrt{5^{2}-4\times 6\left(-4\right)}}{2\times 6}$$

Pūrua 5.

$$y=\frac{-5±\sqrt{25-4\times 6\left(-4\right)}}{2\times 6}$$

Whakareatia -4 ki te 6.

$$y=\frac{-5±\sqrt{25-24\left(-4\right)}}{2\times 6}$$

Whakareatia -24 ki te -4.

$$y=\frac{-5±\sqrt{25+96}}{2\times 6}$$

Tāpiri 25 ki te 96.

$$y=\frac{-5±\sqrt{121}}{2\times 6}$$

Tuhia te pūtakerua o te 121.

$$y=\frac{-5±11}{2\times 6}$$

Whakareatia 2 ki te 6.

$$y=\frac{-5±11}{12}$$

Nā, me whakaoti te whārite y=\frac{-5±11}{12} ina he tāpiri te ±. Tāpiri -5 ki te 11.

$$y=\frac{6}{12}$$

Whakahekea te hautanga \frac{6}{12} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

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

Nā, me whakaoti te whārite y=\frac{-5±11}{12} ina he tango te ±. Tango 11 mai i -5.

$$y=-\frac{16}{12}$$

Whakahekea te hautanga \frac{-16}{12} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$y=-\frac{4}{3}$$

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}{3} mō te x_{2}.

$$6y^{2}+5y-4=6\left(y-\frac{1}{2}\right)\left(y-\left(-\frac{4}{3}\right)\right)$$

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

$$6y^{2}+5y-4=6\left(y-\frac{1}{2}\right)\left(y+\frac{4}{3}\right)$$

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.

$$6y^{2}+5y-4=6\times \frac{2y-1}{2}\left(y+\frac{4}{3}\right)$$

Tāpiri \frac{4}{3} 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.

$$6y^{2}+5y-4=6\times \frac{2y-1}{2}\times \frac{3y+4}{3}$$

Whakareatia \frac{2y-1}{2} ki te \frac{3y+4}{3} 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.

$$6y^{2}+5y-4=6\times \frac{\left(2y-1\right)\left(3y+4\right)}{2\times 3}$$

Whakareatia 2 ki te 3.

$$6y^{2}+5y-4=6\times \frac{\left(2y-1\right)\left(3y+4\right)}{6}$$

Whakakorea atu te tauwehe pūnoa nui rawa 6 i roto i te 6 me te 6.

$$6y^{2}+5y-4=\left(2y-1\right)\left(3y+4\right)$$

Aromātai

$$\left(2y-1\right)\left(3y+4\right)$$