Whakaoti i te sqrt{-x+12}=x

$$\sqrt { - x + 12 } = x$$

Whakaoti mō x

$$x=3$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

$$\left(\sqrt{-x+12}\right)^{2}=x^{2}$$

Tātaihia te \sqrt{-x+12} mā te pū o 2, kia riro ko -x+12.

$$-x+12=x^{2}$$

Tangohia te x^{2} mai i ngā taha e rua.

$$-x+12-x^{2}=0$$

Hurinahatia te pūrau ki te āhua tānga ngahuru. Whakaraupapahia ngā kīanga tau mai i te pū teitei rawa ki te mea iti rawa.

$$-x^{2}-x+12=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei -x^{2}+ax+bx+12. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-1 ab=-12=-12$$

I te mea kua tōraro te ab, he tauaro ngā tohu o a me b. I te mea kua tōraro te a+b, he nui ake te uara pū o te tau tōraro i tō te tōrunga. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua -12.

$$1,-12 2,-6 3,-4$$

Tātaihia te tapeke mō ia takirua.

$$1-12=-11 2-6=-4 3-4=-1$$

Ko te otinga te takirua ka hoatu i te tapeke -1.

$$a=3 b=-4$$

Tuhia anō te -x^{2}-x+12 hei \left(-x^{2}+3x\right)+\left(-4x+12\right).

$$\left(-x^{2}+3x\right)+\left(-4x+12\right)$$

Tauwehea te x i te tuatahi me te 4 i te rōpū tuarua.

$$x\left(-x+3\right)+4\left(-x+3\right)$$

Whakatauwehea atu te kīanga pātahi -x+3 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(-x+3\right)\left(x+4\right)$$

Hei kimi otinga whārite, me whakaoti te -x+3=0 me te x+4=0.

$$x=3 x=-4$$

Whakakapia te 3 mō te x i te whārite \sqrt{-x+12}=x.

$$\sqrt{-3+12}=3$$

Whakarūnātia. Ko te uara x=3 kua ngata te whārite.

$$3=3$$

Whakakapia te -4 mō te x i te whārite \sqrt{-x+12}=x.

$$\sqrt{-\left(-4\right)+12}=-4$$

Whakarūnātia. Ko te uara x=-4 kāore e ngata ana ki te whārite nā te mea e rerekē ngā tohu o te taha maui me te taha katau.

$$4=-4$$

Ko te whārite \sqrt{12-x}=x he rongoā ahurei.

$$x=3$$