Whakaoti i te 1*a^2+11a+10=

$$1 \cdot a ^ { 2 } + 11 a + 10 =$$

Tauwehe

$$\left(a+1\right)\left(a+10\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei a^{2}+pa+qa+10. Hei kimi p me q, whakaritea tētahi pūnaha kia whakaoti.

$$p+q=11 pq=1\times 10=10$$

I te mea kua tōrunga te pq, he ōrite te tohu o p me q. I te mea kua tōrunga te p+q, he tōrunga hoki a p me q. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 10.

$$1,10 2,5$$

Tātaihia te tapeke mō ia takirua.

$$1+10=11 2+5=7$$

Ko te otinga te takirua ka hoatu i te tapeke 11.

$$p=1 q=10$$

Tuhia anō te a^{2}+11a+10 hei \left(a^{2}+a\right)+\left(10a+10\right).

$$\left(a^{2}+a\right)+\left(10a+10\right)$$

Tauwehea te a i te tuatahi me te 10 i te rōpū tuarua.

$$a\left(a+1\right)+10\left(a+1\right)$$

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

$$\left(a+1\right)\left(a+10\right)$$

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.

$$a^{2}+11a+10=0$$

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.

$$a=\frac{-11±\sqrt{11^{2}-4\times 10}}{2}$$

Pūrua 11.

$$a=\frac{-11±\sqrt{121-4\times 10}}{2}$$

Whakareatia -4 ki te 10.

$$a=\frac{-11±\sqrt{121-40}}{2}$$

Tāpiri 121 ki te -40.

$$a=\frac{-11±\sqrt{81}}{2}$$

Tuhia te pūtakerua o te 81.

$$a=\frac{-11±9}{2}$$

Nā, me whakaoti te whārite a=\frac{-11±9}{2} ina he tāpiri te ±. Tāpiri -11 ki te 9.

$$a=-\frac{2}{2}$$

Whakawehe -2 ki te 2.

$$a=-1$$

Nā, me whakaoti te whārite a=\frac{-11±9}{2} ina he tango te ±. Tango 9 mai i -11.

$$a=-\frac{20}{2}$$

Whakawehe -20 ki te 2.

$$a=-10$$

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 -1 mō te x_{1} me te -10 mō te x_{2}.

$$a^{2}+11a+10=\left(a-\left(-1\right)\right)\left(a-\left(-10\right)\right)$$

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

$$a^{2}+11a+10=\left(a+1\right)\left(a+10\right)$$

Aromātai

$$\left(a+1\right)\left(a+10\right)$$