Whakaoti i te 10s^2+19s-15

$$10 s ^ { 2 } + 19 s - 15$$

Tauwehe

$$\left(5s-3\right)\left(2s+5\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 10s^{2}+as+bs-15. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=19 ab=10\left(-15\right)=-150$$

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 -150.

$$-1,150 -2,75 -3,50 -5,30 -6,25 -10,15$$

Tātaihia te tapeke mō ia takirua.

$$-1+150=149 -2+75=73 -3+50=47 -5+30=25 -6+25=19 -10+15=5$$

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

$$a=-6 b=25$$

Tuhia anō te 10s^{2}+19s-15 hei \left(10s^{2}-6s\right)+\left(25s-15\right).

$$\left(10s^{2}-6s\right)+\left(25s-15\right)$$

Tauwehea te 2s i te tuatahi me te 5 i te rōpū tuarua.

$$2s\left(5s-3\right)+5\left(5s-3\right)$$

Whakatauwehea atu te kīanga pātahi 5s-3 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(5s-3\right)\left(2s+5\right)$$

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.

$$10s^{2}+19s-15=0$$

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.

$$s=\frac{-19±\sqrt{19^{2}-4\times 10\left(-15\right)}}{2\times 10}$$

Pūrua 19.

$$s=\frac{-19±\sqrt{361-4\times 10\left(-15\right)}}{2\times 10}$$

Whakareatia -4 ki te 10.

$$s=\frac{-19±\sqrt{361-40\left(-15\right)}}{2\times 10}$$

Whakareatia -40 ki te -15.

$$s=\frac{-19±\sqrt{361+600}}{2\times 10}$$

Tāpiri 361 ki te 600.

$$s=\frac{-19±\sqrt{961}}{2\times 10}$$

Tuhia te pūtakerua o te 961.

$$s=\frac{-19±31}{2\times 10}$$

Whakareatia 2 ki te 10.

$$s=\frac{-19±31}{20}$$

Nā, me whakaoti te whārite s=\frac{-19±31}{20} ina he tāpiri te ±. Tāpiri -19 ki te 31.

$$s=\frac{12}{20}$$

Whakahekea te hautanga \frac{12}{20} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$s=\frac{3}{5}$$

Nā, me whakaoti te whārite s=\frac{-19±31}{20} ina he tango te ±. Tango 31 mai i -19.

$$s=-\frac{50}{20}$$

Whakahekea te hautanga \frac{-50}{20} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 10.

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

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{3}{5} mō te x_{1} me te -\frac{5}{2} mō te x_{2}.

$$10s^{2}+19s-15=10\left(s-\frac{3}{5}\right)\left(s-\left(-\frac{5}{2}\right)\right)$$

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

$$10s^{2}+19s-15=10\left(s-\frac{3}{5}\right)\left(s+\frac{5}{2}\right)$$

Tango \frac{3}{5} mai i s 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.

$$10s^{2}+19s-15=10\times \frac{5s-3}{5}\left(s+\frac{5}{2}\right)$$

Tāpiri \frac{5}{2} ki te s 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.

$$10s^{2}+19s-15=10\times \frac{5s-3}{5}\times \frac{2s+5}{2}$$

Whakareatia \frac{5s-3}{5} ki te \frac{2s+5}{2} 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.

$$10s^{2}+19s-15=10\times \frac{\left(5s-3\right)\left(2s+5\right)}{5\times 2}$$

Whakareatia 5 ki te 2.

$$10s^{2}+19s-15=10\times \frac{\left(5s-3\right)\left(2s+5\right)}{10}$$

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

$$10s^{2}+19s-15=\left(5s-3\right)\left(2s+5\right)$$

Aromātai

$$\left(5s-3\right)\left(2s+5\right)$$