Whakaoti i te 6x^2-19x+10

$$6 { x }^{ 2 } -19x+10$$

Tauwehe

$$\left(2x-5\right)\left(3x-2\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 6x^{2}+ax+bx+10. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-19 ab=6\times 10=60$$

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 60.

$$-1,-60 -2,-30 -3,-20 -4,-15 -5,-12 -6,-10$$

Tātaihia te tapeke mō ia takirua.

$$-1-60=-61 -2-30=-32 -3-20=-23 -4-15=-19 -5-12=-17 -6-10=-16$$

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

$$a=-15 b=-4$$

Tuhia anō te 6x^{2}-19x+10 hei \left(6x^{2}-15x\right)+\left(-4x+10\right).

$$\left(6x^{2}-15x\right)+\left(-4x+10\right)$$

Tauwehea te 3x i te tuatahi me te -2 i te rōpū tuarua.

$$3x\left(2x-5\right)-2\left(2x-5\right)$$

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

$$\left(2x-5\right)\left(3x-2\right)$$

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.

$$6x^{2}-19x+10=0$$

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.

$$x=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}-4\times 6\times 10}}{2\times 6}$$

Pūrua -19.

$$x=\frac{-\left(-19\right)±\sqrt{361-4\times 6\times 10}}{2\times 6}$$

Whakareatia -4 ki te 6.

$$x=\frac{-\left(-19\right)±\sqrt{361-24\times 10}}{2\times 6}$$

Whakareatia -24 ki te 10.

$$x=\frac{-\left(-19\right)±\sqrt{361-240}}{2\times 6}$$

Tāpiri 361 ki te -240.

$$x=\frac{-\left(-19\right)±\sqrt{121}}{2\times 6}$$

Tuhia te pūtakerua o te 121.

$$x=\frac{-\left(-19\right)±11}{2\times 6}$$

Ko te tauaro o -19 ko 19.

$$x=\frac{19±11}{2\times 6}$$

Whakareatia 2 ki te 6.

$$x=\frac{19±11}{12}$$

Nā, me whakaoti te whārite x=\frac{19±11}{12} ina he tāpiri te ±. Tāpiri 19 ki te 11.

$$x=\frac{30}{12}$$

Whakahekea te hautanga \frac{30}{12} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

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

Nā, me whakaoti te whārite x=\frac{19±11}{12} ina he tango te ±. Tango 11 mai i 19.

$$x=\frac{8}{12}$$

Whakahekea te hautanga \frac{8}{12} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$x=\frac{2}{3}$$

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

$$6x^{2}-19x+10=6\left(x-\frac{5}{2}\right)\left(x-\frac{2}{3}\right)$$

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

$$6x^{2}-19x+10=6\times \frac{2x-5}{2}\left(x-\frac{2}{3}\right)$$

Tango \frac{2}{3} mai i x 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.

$$6x^{2}-19x+10=6\times \frac{2x-5}{2}\times \frac{3x-2}{3}$$

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

$$6x^{2}-19x+10=6\times \frac{\left(2x-5\right)\left(3x-2\right)}{2\times 3}$$

Whakareatia 2 ki te 3.

$$6x^{2}-19x+10=6\times \frac{\left(2x-5\right)\left(3x-2\right)}{6}$$

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

$$6x^{2}-19x+10=\left(2x-5\right)\left(3x-2\right)$$

Aromātai

$$\left(2x-5\right)\left(3x-2\right)$$