Whakaoti i te 12k^2+16k-3

$$12 k ^ { 2 } + 16 k - 3$$

Tauwehe

$$\left(6k-1\right)\left(2k+3\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 12k^{2}+ak+bk-3. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=16 ab=12\left(-3\right)=-36$$

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 -36.

$$-1,36 -2,18 -3,12 -4,9 -6,6$$

Tātaihia te tapeke mō ia takirua.

$$-1+36=35 -2+18=16 -3+12=9 -4+9=5 -6+6=0$$

Ko te otinga te takirua ka hoatu i te tapeke 16.

$$a=-2 b=18$$

Tuhia anō te 12k^{2}+16k-3 hei \left(12k^{2}-2k\right)+\left(18k-3\right).

$$\left(12k^{2}-2k\right)+\left(18k-3\right)$$

Tauwehea te 2k i te tuatahi me te 3 i te rōpū tuarua.

$$2k\left(6k-1\right)+3\left(6k-1\right)$$

Whakatauwehea atu te kīanga pātahi 6k-1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(6k-1\right)\left(2k+3\right)$$

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.

$$12k^{2}+16k-3=0$$

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.

$$k=\frac{-16±\sqrt{16^{2}-4\times 12\left(-3\right)}}{2\times 12}$$

Pūrua 16.

$$k=\frac{-16±\sqrt{256-4\times 12\left(-3\right)}}{2\times 12}$$

Whakareatia -4 ki te 12.

$$k=\frac{-16±\sqrt{256-48\left(-3\right)}}{2\times 12}$$

Whakareatia -48 ki te -3.

$$k=\frac{-16±\sqrt{256+144}}{2\times 12}$$

Tāpiri 256 ki te 144.

$$k=\frac{-16±\sqrt{400}}{2\times 12}$$

Tuhia te pūtakerua o te 400.

$$k=\frac{-16±20}{2\times 12}$$

Whakareatia 2 ki te 12.

$$k=\frac{-16±20}{24}$$

Nā, me whakaoti te whārite k=\frac{-16±20}{24} ina he tāpiri te ±. Tāpiri -16 ki te 20.

$$k=\frac{4}{24}$$

Whakahekea te hautanga \frac{4}{24} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$k=\frac{1}{6}$$

Nā, me whakaoti te whārite k=\frac{-16±20}{24} ina he tango te ±. Tango 20 mai i -16.

$$k=-\frac{36}{24}$$

Whakahekea te hautanga \frac{-36}{24} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 12.

$$k=-\frac{3}{2}$$

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{1}{6} mō te x_{1} me te -\frac{3}{2} mō te x_{2}.

$$12k^{2}+16k-3=12\left(k-\frac{1}{6}\right)\left(k-\left(-\frac{3}{2}\right)\right)$$

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

$$12k^{2}+16k-3=12\left(k-\frac{1}{6}\right)\left(k+\frac{3}{2}\right)$$

Tango \frac{1}{6} mai i k 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.

$$12k^{2}+16k-3=12\times \frac{6k-1}{6}\left(k+\frac{3}{2}\right)$$

Tāpiri \frac{3}{2} ki te k 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.

$$12k^{2}+16k-3=12\times \frac{6k-1}{6}\times \frac{2k+3}{2}$$

Whakareatia \frac{6k-1}{6} ki te \frac{2k+3}{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.

$$12k^{2}+16k-3=12\times \frac{\left(6k-1\right)\left(2k+3\right)}{6\times 2}$$

Whakareatia 6 ki te 2.

$$12k^{2}+16k-3=12\times \frac{\left(6k-1\right)\left(2k+3\right)}{12}$$

Whakakorea atu te tauwehe pūnoa nui rawa 12 i roto i te 12 me te 12.

$$12k^{2}+16k-3=\left(6k-1\right)\left(2k+3\right)$$

Aromātai

$$\left(6k-1\right)\left(2k+3\right)$$