Whakaoti i te 8x^2+10x-3

$$8 x ^ { 2 } + 10 x - 3$$

Tauwehe

$$\left(4x-1\right)\left(2x+3\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 8x^{2}+ax+bx-3. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=10 ab=8\left(-3\right)=-24$$

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 -24.

$$-1,24 -2,12 -3,8 -4,6$$

Tātaihia te tapeke mō ia takirua.

$$-1+24=23 -2+12=10 -3+8=5 -4+6=2$$

Ko te otinga te takirua ka hoatu i te tapeke 10.

$$a=-2 b=12$$

Tuhia anō te 8x^{2}+10x-3 hei \left(8x^{2}-2x\right)+\left(12x-3\right).

$$\left(8x^{2}-2x\right)+\left(12x-3\right)$$

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

$$2x\left(4x-1\right)+3\left(4x-1\right)$$

Whakatauwehea atu te kīanga pātahi 4x-1 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(4x-1\right)\left(2x+3\right)$$

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.

$$8x^{2}+10x-3=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\times 8\left(-3\right)}}{2\times 8}$$

Pūrua 10.

$$x=\frac{-10±\sqrt{100-4\times 8\left(-3\right)}}{2\times 8}$$

Whakareatia -4 ki te 8.

$$x=\frac{-10±\sqrt{100-32\left(-3\right)}}{2\times 8}$$

Whakareatia -32 ki te -3.

$$x=\frac{-10±\sqrt{100+96}}{2\times 8}$$

Tāpiri 100 ki te 96.

$$x=\frac{-10±\sqrt{196}}{2\times 8}$$

Tuhia te pūtakerua o te 196.

$$x=\frac{-10±14}{2\times 8}$$

Whakareatia 2 ki te 8.

$$x=\frac{-10±14}{16}$$

Nā, me whakaoti te whārite x=\frac{-10±14}{16} ina he tāpiri te ±. Tāpiri -10 ki te 14.

$$x=\frac{4}{16}$$

Whakahekea te hautanga \frac{4}{16} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

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

Nā, me whakaoti te whārite x=\frac{-10±14}{16} ina he tango te ±. Tango 14 mai i -10.

$$x=-\frac{24}{16}$$

Whakahekea te hautanga \frac{-24}{16} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 8.

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

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

$$8x^{2}+10x-3=8\left(x-\frac{1}{4}\right)\left(x-\left(-\frac{3}{2}\right)\right)$$

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

$$8x^{2}+10x-3=8\left(x-\frac{1}{4}\right)\left(x+\frac{3}{2}\right)$$

Tango \frac{1}{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.

$$8x^{2}+10x-3=8\times \frac{4x-1}{4}\left(x+\frac{3}{2}\right)$$

Tāpiri \frac{3}{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.

$$8x^{2}+10x-3=8\times \frac{4x-1}{4}\times \frac{2x+3}{2}$$

Whakareatia \frac{4x-1}{4} ki te \frac{2x+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.

$$8x^{2}+10x-3=8\times \frac{\left(4x-1\right)\left(2x+3\right)}{4\times 2}$$

Whakareatia 4 ki te 2.

$$8x^{2}+10x-3=8\times \frac{\left(4x-1\right)\left(2x+3\right)}{8}$$

Whakakorea atu te tauwehe pūnoa nui rawa 8 i roto i te 8 me te 8.

$$8x^{2}+10x-3=\left(4x-1\right)\left(2x+3\right)$$

Aromātai

$$\left(4x-1\right)\left(2x+3\right)$$