Whakaoti i te 2k^2-5k-18=

$$2 k ^ { 2 } - 5 k - 18 =$$

Tauwehe

$$\left(2k-9\right)\left(k+2\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 2k^{2}+ak+bk-18. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-5 ab=2\left(-18\right)=-36$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. 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 -5.

$$a=-9 b=4$$

Tuhia anō te 2k^{2}-5k-18 hei \left(2k^{2}-9k\right)+\left(4k-18\right).

$$\left(2k^{2}-9k\right)+\left(4k-18\right)$$

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

$$k\left(2k-9\right)+2\left(2k-9\right)$$

Whakatauwehea atu te kīanga pātahi 2k-9 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2k-9\right)\left(k+2\right)$$

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.

$$2k^{2}-5k-18=0$$

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.

$$k=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 2\left(-18\right)}}{2\times 2}$$

Pūrua -5.

$$k=\frac{-\left(-5\right)±\sqrt{25-4\times 2\left(-18\right)}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$k=\frac{-\left(-5\right)±\sqrt{25-8\left(-18\right)}}{2\times 2}$$

Whakareatia -8 ki te -18.

$$k=\frac{-\left(-5\right)±\sqrt{25+144}}{2\times 2}$$

Tāpiri 25 ki te 144.

$$k=\frac{-\left(-5\right)±\sqrt{169}}{2\times 2}$$

Tuhia te pūtakerua o te 169.

$$k=\frac{-\left(-5\right)±13}{2\times 2}$$

Ko te tauaro o -5 ko 5.

$$k=\frac{5±13}{2\times 2}$$

Whakareatia 2 ki te 2.

$$k=\frac{5±13}{4}$$

Nā, me whakaoti te whārite k=\frac{5±13}{4} ina he tāpiri te ±. Tāpiri 5 ki te 13.

$$k=\frac{18}{4}$$

Whakahekea te hautanga \frac{18}{4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$k=\frac{9}{2}$$

Nā, me whakaoti te whārite k=\frac{5±13}{4} ina he tango te ±. Tango 13 mai i 5.

$$k=-\frac{8}{4}$$

Whakawehe -8 ki te 4.

$$k=-2$$

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{9}{2} mō te x_{1} me te -2 mō te x_{2}.

$$2k^{2}-5k-18=2\left(k-\frac{9}{2}\right)\left(k-\left(-2\right)\right)$$

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

$$2k^{2}-5k-18=2\left(k-\frac{9}{2}\right)\left(k+2\right)$$

Tango \frac{9}{2} 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.

$$2k^{2}-5k-18=2\times \frac{2k-9}{2}\left(k+2\right)$$

Whakakorea atu te tauwehe pūnoa nui rawa 2 i roto i te 2 me te 2.

$$2k^{2}-5k-18=\left(2k-9\right)\left(k+2\right)$$

Aromātai

$$\left(2k-9\right)\left(k+2\right)$$