Whakaoti i te x^2+4x-5

$$x ^ { 2 } + 4 x - 5$$

Tauwehe

$$\left(x-1\right)\left(x+5\right)$$

Ngā Upane Otinga

Whakatauwehea te kīanga mā te whakarōpū. Tuatahi, me tuhi anō te kīanga hei x^{2}+ax+bx-5. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=4 ab=1\left(-5\right)=-5$$

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. Ko te takirua anake pērā ko te otinga pūnaha.

$$a=-1 b=5$$

Tuhia anō te x^{2}+4x-5 hei \left(x^{2}-x\right)+\left(5x-5\right).

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

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

$$x\left(x-1\right)+5\left(x-1\right)$$

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

$$\left(x-1\right)\left(x+5\right)$$

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.

$$x^{2}+4x-5=0$$

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.

$$x=\frac{-4±\sqrt{4^{2}-4\left(-5\right)}}{2}$$

Pūrua 4.

$$x=\frac{-4±\sqrt{16-4\left(-5\right)}}{2}$$

Whakareatia -4 ki te -5.

$$x=\frac{-4±\sqrt{16+20}}{2}$$

Tāpiri 16 ki te 20.

$$x=\frac{-4±\sqrt{36}}{2}$$

Tuhia te pūtakerua o te 36.

$$x=\frac{-4±6}{2}$$

Nā, me whakaoti te whārite x=\frac{-4±6}{2} ina he tāpiri te ±. Tāpiri -4 ki te 6.

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

Whakawehe 2 ki te 2.

$$x=1$$

Nā, me whakaoti te whārite x=\frac{-4±6}{2} ina he tango te ±. Tango 6 mai i -4.

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

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 1 mō te x_{1} me te -5 mō te x_{2}.

$$x^{2}+4x-5=\left(x-1\right)\left(x-\left(-5\right)\right)$$

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

$$x^{2}+4x-5=\left(x-1\right)\left(x+5\right)$$

Aromātai

$$\left(x-1\right)\left(x+5\right)$$