Whakaoti i te 4t^2+4t-3

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

Tauwehe

$$\left(2t-1\right)\left(2t+3\right)$$

Ngā Upane Otinga

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

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

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 4.

$$a=-2 b=6$$

Tuhia anō te 4t^{2}+4t-3 hei \left(4t^{2}-2t\right)+\left(6t-3\right).

$$\left(4t^{2}-2t\right)+\left(6t-3\right)$$

Tauwehea te 2t i te tuatahi me te 3 i te rōpū tuarua.

$$2t\left(2t-1\right)+3\left(2t-1\right)$$

Whakatauwehea atu te kīanga pātahi 2t-1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2t-1\right)\left(2t+3\right)$$

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.

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

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.

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

Pūrua 4.

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

Whakareatia -4 ki te 4.

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

Whakareatia -16 ki te -3.

$$t=\frac{-4±\sqrt{16+48}}{2\times 4}$$

Tāpiri 16 ki te 48.

$$t=\frac{-4±\sqrt{64}}{2\times 4}$$

Tuhia te pūtakerua o te 64.

$$t=\frac{-4±8}{2\times 4}$$

Whakareatia 2 ki te 4.

$$t=\frac{-4±8}{8}$$

Nā, me whakaoti te whārite t=\frac{-4±8}{8} ina he tāpiri te ±. Tāpiri -4 ki te 8.

$$t=\frac{4}{8}$$

Whakahekea te hautanga \frac{4}{8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$t=\frac{1}{2}$$

Nā, me whakaoti te whārite t=\frac{-4±8}{8} ina he tango te ±. Tango 8 mai i -4.

$$t=-\frac{12}{8}$$

Whakahekea te hautanga \frac{-12}{8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$t=-\frac{3}{2}$$

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

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

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

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

Tango \frac{1}{2} mai i t 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.

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

Tāpiri \frac{3}{2} ki te t 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.

$$4t^{2}+4t-3=4\times \frac{2t-1}{2}\times \frac{2t+3}{2}$$

Whakareatia \frac{2t-1}{2} ki te \frac{2t+3}{2} mā te whakarea taurunga ki te taurunga me te tauraro ki te tauraro, ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

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

Whakareatia 2 ki te 2.

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

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

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

Aromātai

$$\left(2t-1\right)\left(2t+3\right)$$