Whakaoti i te 4x^2+x-33

$$4 x ^ { 2 } + x - 33$$

Tauwehe

$$\left(4x-11\right)\left(x+3\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 4x^{2}+ax+bx-33. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=1 ab=4\left(-33\right)=-132$$

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 -132.

$$-1,132 -2,66 -3,44 -4,33 -6,22 -11,12$$

Tātaihia te tapeke mō ia takirua.

$$-1+132=131 -2+66=64 -3+44=41 -4+33=29 -6+22=16 -11+12=1$$

Ko te otinga te takirua ka hoatu i te tapeke 1.

$$a=-11 b=12$$

Tuhia anō te 4x^{2}+x-33 hei \left(4x^{2}-11x\right)+\left(12x-33\right).

$$\left(4x^{2}-11x\right)+\left(12x-33\right)$$

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

$$x\left(4x-11\right)+3\left(4x-11\right)$$

Whakatauwehea atu te kīanga pātahi 4x-11 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(4x-11\right)\left(x+3\right)$$

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.

$$4x^{2}+x-33=0$$

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.

$$x=\frac{-1±\sqrt{1^{2}-4\times 4\left(-33\right)}}{2\times 4}$$

Pūrua 1.

$$x=\frac{-1±\sqrt{1-4\times 4\left(-33\right)}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$x=\frac{-1±\sqrt{1-16\left(-33\right)}}{2\times 4}$$

Whakareatia -16 ki te -33.

$$x=\frac{-1±\sqrt{1+528}}{2\times 4}$$

Tāpiri 1 ki te 528.

$$x=\frac{-1±\sqrt{529}}{2\times 4}$$

Tuhia te pūtakerua o te 529.

$$x=\frac{-1±23}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{-1±23}{8}$$

Nā, me whakaoti te whārite x=\frac{-1±23}{8} ina he tāpiri te ±. Tāpiri -1 ki te 23.

$$x=\frac{22}{8}$$

Whakahekea te hautanga \frac{22}{8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=\frac{11}{4}$$

Nā, me whakaoti te whārite x=\frac{-1±23}{8} ina he tango te ±. Tango 23 mai i -1.

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

Whakawehe -24 ki te 8.

$$x=-3$$

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{11}{4} mō te x_{1} me te -3 mō te x_{2}.

$$4x^{2}+x-33=4\left(x-\frac{11}{4}\right)\left(x-\left(-3\right)\right)$$

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

$$4x^{2}+x-33=4\left(x-\frac{11}{4}\right)\left(x+3\right)$$

Tango \frac{11}{4} 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.

$$4x^{2}+x-33=4\times \frac{4x-11}{4}\left(x+3\right)$$

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

$$4x^{2}+x-33=\left(4x-11\right)\left(x+3\right)$$

Aromātai

$$\left(4x-11\right)\left(x+3\right)$$