Whakaoti i te =x^2+11x+10=0

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

Whakaoti mō x

$$x=-10$$

$$x=-1$$

Ngā Upane Whakamahi Tauwehe

Hei whakaoti i te whārite, whakatauwehea te x^{2}+11x+10 mā te whakamahi i te tātai x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=11 ab=10$$

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 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.

$$a=1 b=10$$

Me tuhi anō te kīanga whakatauwehe \left(x+a\right)\left(x+b\right) mā ngā uara i tātaihia.

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

Hei kimi otinga whārite, me whakaoti te x+1=0 me te x+10=0.

$$x=-1 x=-10$$

Ngā Upane Whakamahi Tauwehe Mā te Whakarōpū

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei x^{2}+ax+bx+10. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=11 ab=1\times 10=10$$

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 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.

$$a=1 b=10$$

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

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

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

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

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

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

Hei kimi otinga whārite, me whakaoti te x+1=0 me te x+10=0.

$$x=-1 x=-10$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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.

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

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 11 mō b, me 10 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Pūrua 11.

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

Whakareatia -4 ki te 10.

$$x=\frac{-11±\sqrt{121-40}}{2}$$

Tāpiri 121 ki te -40.

$$x=\frac{-11±\sqrt{81}}{2}$$

Tuhia te pūtakerua o te 81.

$$x=\frac{-11±9}{2}$$

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

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

Whakawehe -2 ki te 2.

$$x=-1$$

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

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

Whakawehe -20 ki te 2.

$$x=-10$$

Kua oti te whārite te whakatau.

$$x=-1 x=-10$$

Ngā Upane mō te Whakaoti i te Pūrua

Ko ngā whārite pūrua pēnei i tēnei nā ka taea te whakaoti mā te whakaoti i te pūrua. Hei whakaoti i te pūrua, ko te whārite me mātua tuhi ki te āhua x^{2}+bx=c.

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

Me tango 10 mai i ngā taha e rua o te whārite.

$$x^{2}+11x+10-10=-10$$

Mā te tango i te 10 i a ia ake anō ka toe ko te 0.

$$x^{2}+11x=-10$$

Whakawehea te 11, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{11}{2}. Nā, tāpiria te pūrua o te \frac{11}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+11x+\left(\frac{11}{2}\right)^{2}=-10+\left(\frac{11}{2}\right)^{2}$$

Pūruatia \frac{11}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+11x+\frac{121}{4}=-10+\frac{121}{4}$$

Tāpiri -10 ki te \frac{121}{4}.

$$x^{2}+11x+\frac{121}{4}=\frac{81}{4}$$

Tauwehea x^{2}+11x+\frac{121}{4}. Ko te tikanga pūnoa, ina ko x^{2}+bx+c he pūrua tika pūrua tika pū, ka taea taua mea te tauwehea i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{11}{2}\right)^{2}=\frac{81}{4}$$

Tuhia te pūtakerua o ngā taha e rua o te whārite.

$$\sqrt{\left(x+\frac{11}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Whakarūnātia.

$$x+\frac{11}{2}=\frac{9}{2} x+\frac{11}{2}=-\frac{9}{2}$$

Me tango \frac{11}{2} mai i ngā taha e rua o te whārite.

$$x=-1 x=-10$$