Whakaoti i te 4x^2-11x-3

$$4 { x }^{ 2 } -11x-3$$

Tauwehe

$$\left(x-3\right)\left(4x+1\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 4x^{2}+ax+bx-3. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-11 ab=4\left(-3\right)=-12$$

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 -12.

$$1,-12 2,-6 3,-4$$

Tātaihia te tapeke mō ia takirua.

$$1-12=-11 2-6=-4 3-4=-1$$

Ko te otinga te takirua ka hoatu i te tapeke -11.

$$a=-12 b=1$$

Tuhia anō te 4x^{2}-11x-3 hei \left(4x^{2}-12x\right)+\left(x-3\right).

$$\left(4x^{2}-12x\right)+\left(x-3\right)$$

Whakatauwehea atu 4x i te 4x^{2}-12x.

$$4x\left(x-3\right)+x-3$$

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

$$\left(x-3\right)\left(4x+1\right)$$

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.

$$4x^{2}-11x-3=0$$

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.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 4\left(-3\right)}}{2\times 4}$$

Pūrua -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\times 4\left(-3\right)}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$x=\frac{-\left(-11\right)±\sqrt{121-16\left(-3\right)}}{2\times 4}$$

Whakareatia -16 ki te -3.

$$x=\frac{-\left(-11\right)±\sqrt{121+48}}{2\times 4}$$

Tāpiri 121 ki te 48.

$$x=\frac{-\left(-11\right)±\sqrt{169}}{2\times 4}$$

Tuhia te pūtakerua o te 169.

$$x=\frac{-\left(-11\right)±13}{2\times 4}$$

Ko te tauaro o -11 ko 11.

$$x=\frac{11±13}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{11±13}{8}$$

Nā, me whakaoti te whārite x=\frac{11±13}{8} ina he tāpiri te ±. Tāpiri 11 ki te 13.

$$x=\frac{24}{8}$$

Whakawehe 24 ki te 8.

$$x=3$$

Nā, me whakaoti te whārite x=\frac{11±13}{8} ina he tango te ±. Tango 13 mai i 11.

$$x=-\frac{2}{8}$$

Whakahekea te hautanga \frac{-2}{8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=-\frac{1}{4}$$

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}{4} mō te x_{2}.

$$4x^{2}-11x-3=4\left(x-3\right)\left(x-\left(-\frac{1}{4}\right)\right)$$

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

$$4x^{2}-11x-3=4\left(x-3\right)\left(x+\frac{1}{4}\right)$$

Tāpiri \frac{1}{4} 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.

$$4x^{2}-11x-3=4\left(x-3\right)\times \frac{4x+1}{4}$$

Whakakorea atu te tauwehe pūnoa nui rawa 4 i roto i te 4 me te 4.

$$4x^{2}-11x-3=\left(x-3\right)\left(4x+1\right)$$

Aromātai

$$\left(x-3\right)\left(4x+1\right)$$