Whakaoti i te x^2-8x+15

$$x ^ { 2 } - 8 x + 15$$

Tauwehe

$$\left(x-5\right)\left(x-3\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei x^{2}+ax+bx+15. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-8 ab=1\times 15=15$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōraro te a+b, he tōraro hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 15.

$$-1,-15 -3,-5$$

Tātaihia te tapeke mō ia takirua.

$$-1-15=-16 -3-5=-8$$

Ko te otinga te takirua ka hoatu i te tapeke -8.

$$a=-5 b=-3$$

Tuhia anō te x^{2}-8x+15 hei \left(x^{2}-5x\right)+\left(-3x+15\right).

$$\left(x^{2}-5x\right)+\left(-3x+15\right)$$

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

$$x\left(x-5\right)-3\left(x-5\right)$$

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

$$\left(x-5\right)\left(x-3\right)$$

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.

$$x^{2}-8x+15=0$$

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.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 15}}{2}$$

Pūrua -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 15}}{2}$$

Whakareatia -4 ki te 15.

$$x=\frac{-\left(-8\right)±\sqrt{64-60}}{2}$$

Tāpiri 64 ki te -60.

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

Tuhia te pūtakerua o te 4.

$$x=\frac{-\left(-8\right)±2}{2}$$

Ko te tauaro o -8 ko 8.

$$x=\frac{8±2}{2}$$

Nā, me whakaoti te whārite x=\frac{8±2}{2} ina he tāpiri te ±. Tāpiri 8 ki te 2.

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

Whakawehe 10 ki te 2.

$$x=5$$

Nā, me whakaoti te whārite x=\frac{8±2}{2} ina he tango te ±. Tango 2 mai i 8.

$$x=\frac{6}{2}$$

Whakawehe 6 ki te 2.

$$x=3$$

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 5 mō te x_{1} me te 3 mō te x_{2}.

$$x^{2}-8x+15=\left(x-5\right)\left(x-3\right)$$

Aromātai

$$\left(x-5\right)\left(x-3\right)$$