Whakaoti i te 4x^2+12x+5

$$4 x ^ { 2 } + 12 x + 5$$

Tauwehe

$$\left(2x+1\right)\left(2x+5\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 4x^{2}+ax+bx+5. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=12 ab=4\times 5=20$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōrunga te a+b, he tōrunga hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 20.

$$1,20 2,10 4,5$$

Tātaihia te tapeke mō ia takirua.

$$1+20=21 2+10=12 4+5=9$$

Ko te otinga te takirua ka hoatu i te tapeke 12.

$$a=2 b=10$$

Tuhia anō te 4x^{2}+12x+5 hei \left(4x^{2}+2x\right)+\left(10x+5\right).

$$\left(4x^{2}+2x\right)+\left(10x+5\right)$$

Tauwehea te 2x i te tuatahi me te 5 i te rōpū tuarua.

$$2x\left(2x+1\right)+5\left(2x+1\right)$$

Whakatauwehea atu te kīanga pātahi 2x+1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(2x+1\right)\left(2x+5\right)$$

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.

$$4x^{2}+12x+5=0$$

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.

$$x=\frac{-12±\sqrt{12^{2}-4\times 4\times 5}}{2\times 4}$$

Pūrua 12.

$$x=\frac{-12±\sqrt{144-4\times 4\times 5}}{2\times 4}$$

Whakareatia -4 ki te 4.

$$x=\frac{-12±\sqrt{144-16\times 5}}{2\times 4}$$

Whakareatia -16 ki te 5.

$$x=\frac{-12±\sqrt{144-80}}{2\times 4}$$

Tāpiri 144 ki te -80.

$$x=\frac{-12±\sqrt{64}}{2\times 4}$$

Tuhia te pūtakerua o te 64.

$$x=\frac{-12±8}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{-12±8}{8}$$

Nā, me whakaoti te whārite x=\frac{-12±8}{8} ina he tāpiri te ±. Tāpiri -12 ki te 8.

$$x=-\frac{4}{8}$$

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

$$x=-\frac{1}{2}$$

Nā, me whakaoti te whārite x=\frac{-12±8}{8} ina he tango te ±. Tango 8 mai i -12.

$$x=-\frac{20}{8}$$

Whakahekea te hautanga \frac{-20}{8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$x=-\frac{5}{2}$$

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{5}{2} mō te x_{2}.

$$4x^{2}+12x+5=4\left(x-\left(-\frac{1}{2}\right)\right)\left(x-\left(-\frac{5}{2}\right)\right)$$

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

$$4x^{2}+12x+5=4\left(x+\frac{1}{2}\right)\left(x+\frac{5}{2}\right)$$

Tāpiri \frac{1}{2} 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.

$$4x^{2}+12x+5=4\times \frac{2x+1}{2}\left(x+\frac{5}{2}\right)$$

Tāpiri \frac{5}{2} 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.

$$4x^{2}+12x+5=4\times \frac{2x+1}{2}\times \frac{2x+5}{2}$$

Whakareatia \frac{2x+1}{2} ki te \frac{2x+5}{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.

$$4x^{2}+12x+5=4\times \frac{\left(2x+1\right)\left(2x+5\right)}{2\times 2}$$

Whakareatia 2 ki te 2.

$$4x^{2}+12x+5=4\times \frac{\left(2x+1\right)\left(2x+5\right)}{4}$$

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

$$4x^{2}+12x+5=\left(2x+1\right)\left(2x+5\right)$$

Aromātai

$$\left(2x+1\right)\left(2x+5\right)$$