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

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

Tauwehe

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

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei x^{2}+ax+bx-10. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=3 ab=1\left(-10\right)=-10$$

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 -10.

$$-1,10 -2,5$$

Tātaihia te tapeke mō ia takirua.

$$-1+10=9 -2+5=3$$

Ko te otinga te takirua ka hoatu i te tapeke 3.

$$a=-2 b=5$$

Tuhia anō te x^{2}+3x-10 hei \left(x^{2}-2x\right)+\left(5x-10\right).

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

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

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

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

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

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.

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

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.

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

Pūrua 3.

$$x=\frac{-3±\sqrt{9-4\left(-10\right)}}{2}$$

Whakareatia -4 ki te -10.

$$x=\frac{-3±\sqrt{9+40}}{2}$$

Tāpiri 9 ki te 40.

$$x=\frac{-3±\sqrt{49}}{2}$$

Tuhia te pūtakerua o te 49.

$$x=\frac{-3±7}{2}$$

Nā, me whakaoti te whārite x=\frac{-3±7}{2} ina he tāpiri te ±. Tāpiri -3 ki te 7.

$$x=\frac{4}{2}$$

Whakawehe 4 ki te 2.

$$x=2$$

Nā, me whakaoti te whārite x=\frac{-3±7}{2} ina he tango te ±. Tango 7 mai i -3.

$$x=-\frac{10}{2}$$

Whakawehe -10 ki te 2.

$$x=-5$$

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 2 mō te x_{1} me te -5 mō te x_{2}.

$$x^{2}+3x-10=\left(x-2\right)\left(x-\left(-5\right)\right)$$

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

$$x^{2}+3x-10=\left(x-2\right)\left(x+5\right)$$

Aromātai

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