Whakaoti i te f(x)=25x^2+100x+99

$$f ( x ) = 25 x ^ { 2 } + 100 x + 99$$

Tauwehe

$$\left(5x+9\right)\left(5x+11\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei 25x^{2}+ax+bx+99. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=100 ab=25\times 99=2475$$

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 2475.

$$1,2475 3,825 5,495 9,275 11,225 15,165 25,99 33,75 45,55$$

Tātaihia te tapeke mō ia takirua.

$$1+2475=2476 3+825=828 5+495=500 9+275=284 11+225=236 15+165=180 25+99=124 33+75=108 45+55=100$$

Ko te otinga te takirua ka hoatu i te tapeke 100.

$$a=45 b=55$$

Tuhia anō te 25x^{2}+100x+99 hei \left(25x^{2}+45x\right)+\left(55x+99\right).

$$\left(25x^{2}+45x\right)+\left(55x+99\right)$$

Tauwehea te 5x i te tuatahi me te 11 i te rōpū tuarua.

$$5x\left(5x+9\right)+11\left(5x+9\right)$$

Whakatauwehea atu te kīanga pātahi 5x+9 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(5x+9\right)\left(5x+11\right)$$

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.

$$25x^{2}+100x+99=0$$

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.

$$x=\frac{-100±\sqrt{100^{2}-4\times 25\times 99}}{2\times 25}$$

Pūrua 100.

$$x=\frac{-100±\sqrt{10000-4\times 25\times 99}}{2\times 25}$$

Whakareatia -4 ki te 25.

$$x=\frac{-100±\sqrt{10000-100\times 99}}{2\times 25}$$

Whakareatia -100 ki te 99.

$$x=\frac{-100±\sqrt{10000-9900}}{2\times 25}$$

Tāpiri 10000 ki te -9900.

$$x=\frac{-100±\sqrt{100}}{2\times 25}$$

Tuhia te pūtakerua o te 100.

$$x=\frac{-100±10}{2\times 25}$$

Whakareatia 2 ki te 25.

$$x=\frac{-100±10}{50}$$

Nā, me whakaoti te whārite x=\frac{-100±10}{50} ina he tāpiri te ±. Tāpiri -100 ki te 10.

$$x=-\frac{90}{50}$$

Whakahekea te hautanga \frac{-90}{50} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 10.

$$x=-\frac{9}{5}$$

Nā, me whakaoti te whārite x=\frac{-100±10}{50} ina he tango te ±. Tango 10 mai i -100.

$$x=-\frac{110}{50}$$

Whakahekea te hautanga \frac{-110}{50} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 10.

$$x=-\frac{11}{5}$$

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{9}{5} mō te x_{1} me te -\frac{11}{5} mō te x_{2}.

$$25x^{2}+100x+99=25\left(x-\left(-\frac{9}{5}\right)\right)\left(x-\left(-\frac{11}{5}\right)\right)$$

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

$$25x^{2}+100x+99=25\left(x+\frac{9}{5}\right)\left(x+\frac{11}{5}\right)$$

Tāpiri \frac{9}{5} 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.

$$25x^{2}+100x+99=25\times \frac{5x+9}{5}\left(x+\frac{11}{5}\right)$$

Tāpiri \frac{11}{5} 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.

$$25x^{2}+100x+99=25\times \frac{5x+9}{5}\times \frac{5x+11}{5}$$

Whakareatia \frac{5x+9}{5} ki te \frac{5x+11}{5} 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.

$$25x^{2}+100x+99=25\times \frac{\left(5x+9\right)\left(5x+11\right)}{5\times 5}$$

Whakareatia 5 ki te 5.

$$25x^{2}+100x+99=25\times \frac{\left(5x+9\right)\left(5x+11\right)}{25}$$

Whakakorea atu te tauwehe pūnoa nui rawa 25 i roto i te 25 me te 25.

$$25x^{2}+100x+99=\left(5x+9\right)\left(5x+11\right)$$

Aromātai

$$\left(5x+9\right)\left(5x+11\right)$$