Whakaoti i te 5x^2-14x-3

$$5 x ^ { 2 } - 14 x - 3$$

Tauwehe

$$\left(x-3\right)\left(5x+1\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 5x^{2}+ax+bx-3. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-14 ab=5\left(-3\right)=-15$$

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 -15.

$$1,-15 3,-5$$

Tātaihia te tapeke mō ia takirua.

$$1-15=-14 3-5=-2$$

Ko te otinga te takirua ka hoatu i te tapeke -14.

$$a=-15 b=1$$

Tuhia anō te 5x^{2}-14x-3 hei \left(5x^{2}-15x\right)+\left(x-3\right).

$$\left(5x^{2}-15x\right)+\left(x-3\right)$$

Whakatauwehea atu 5x i te 5x^{2}-15x.

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

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

$$\left(x-3\right)\left(5x+1\right)$$

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.

$$5x^{2}-14x-3=0$$

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.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 5\left(-3\right)}}{2\times 5}$$

Pūrua -14.

$$x=\frac{-\left(-14\right)±\sqrt{196-4\times 5\left(-3\right)}}{2\times 5}$$

Whakareatia -4 ki te 5.

$$x=\frac{-\left(-14\right)±\sqrt{196-20\left(-3\right)}}{2\times 5}$$

Whakareatia -20 ki te -3.

$$x=\frac{-\left(-14\right)±\sqrt{196+60}}{2\times 5}$$

Tāpiri 196 ki te 60.

$$x=\frac{-\left(-14\right)±\sqrt{256}}{2\times 5}$$

Tuhia te pūtakerua o te 256.

$$x=\frac{-\left(-14\right)±16}{2\times 5}$$

Ko te tauaro o -14 ko 14.

$$x=\frac{14±16}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{14±16}{10}$$

Nā, me whakaoti te whārite x=\frac{14±16}{10} ina he tāpiri te ±. Tāpiri 14 ki te 16.

$$x=\frac{30}{10}$$

Whakawehe 30 ki te 10.

$$x=3$$

Nā, me whakaoti te whārite x=\frac{14±16}{10} ina he tango te ±. Tango 16 mai i 14.

$$x=-\frac{2}{10}$$

Whakahekea te hautanga \frac{-2}{10} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=-\frac{1}{5}$$

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

$$5x^{2}-14x-3=5\left(x-3\right)\left(x-\left(-\frac{1}{5}\right)\right)$$

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

$$5x^{2}-14x-3=5\left(x-3\right)\left(x+\frac{1}{5}\right)$$

Tāpiri \frac{1}{5} ki te x 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.

$$5x^{2}-14x-3=5\left(x-3\right)\times \frac{5x+1}{5}$$

Whakakorea atu te tauwehe pūnoa nui rawa 5 i roto i te 5 me te 5.

$$5x^{2}-14x-3=\left(x-3\right)\left(5x+1\right)$$

Aromātai

$$\left(x-3\right)\left(5x+1\right)$$