Whakaoti i te 4u^2+u-3

$$4 u ^ { 2 } + u - 3$$

Tauwehe

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

Ngā Upane Otinga

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

$$a+b=1 ab=4\left(-3\right)=-12$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua 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 -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 1.

$$a=-3 b=4$$

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

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

Whakatauwehea atu u i te 4u^{2}-3u.

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

Whakatauwehea atu te kīanga pātahi 4u-3 mā te whakamahi i te āhuatanga tātai tohatoha.

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

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.

$$4u^{2}+u-3=0$$

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.

$$u=\frac{-1±\sqrt{1^{2}-4\times 4\left(-3\right)}}{2\times 4}$$

Pūrua 1.

$$u=\frac{-1±\sqrt{1-4\times 4\left(-3\right)}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$u=\frac{-1±\sqrt{1-16\left(-3\right)}}{2\times 4}$$

Whakareatia -16 ki te -3.

$$u=\frac{-1±\sqrt{1+48}}{2\times 4}$$

Tāpiri 1 ki te 48.

$$u=\frac{-1±\sqrt{49}}{2\times 4}$$

Tuhia te pūtakerua o te 49.

$$u=\frac{-1±7}{2\times 4}$$

Whakareatia 2 ki te 4.

$$u=\frac{-1±7}{8}$$

Nā, me whakaoti te whārite u=\frac{-1±7}{8} ina he tāpiri te ±. Tāpiri -1 ki te 7.

$$u=\frac{6}{8}$$

Whakahekea te hautanga \frac{6}{8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$u=\frac{3}{4}$$

Nā, me whakaoti te whārite u=\frac{-1±7}{8} ina he tango te ±. Tango 7 mai i -1.

$$u=-\frac{8}{8}$$

Whakawehe -8 ki te 8.

$$u=-1$$

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

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

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

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

Tango \frac{3}{4} mai i u 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.

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

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

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

Aromātai

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