Whakaoti i te sqrt{40-3x}=x

$$\sqrt { 40 - 3 x } = x$$

Whakaoti mō x

$$x=5$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

$$\left(\sqrt{40-3x}\right)^{2}=x^{2}$$

Tātaihia te \sqrt{40-3x} mā te pū o 2, kia riro ko 40-3x.

$$40-3x=x^{2}$$

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

$$40-3x-x^{2}=0$$

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

$$-x^{2}-3x+40=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei -x^{2}+ax+bx+40. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=-3 ab=-40=-40$$

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 -40.

$$1,-40 2,-20 4,-10 5,-8$$

Tātaihia te tapeke mō ia takirua.

$$1-40=-39 2-20=-18 4-10=-6 5-8=-3$$

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

$$a=5 b=-8$$

Tuhia anō te -x^{2}-3x+40 hei \left(-x^{2}+5x\right)+\left(-8x+40\right).

$$\left(-x^{2}+5x\right)+\left(-8x+40\right)$$

Tauwehea te x i te tuatahi me te 8 i te rōpū tuarua.

$$x\left(-x+5\right)+8\left(-x+5\right)$$

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

$$\left(-x+5\right)\left(x+8\right)$$

Hei kimi otinga whārite, me whakaoti te -x+5=0 me te x+8=0.

$$x=5 x=-8$$

Whakakapia te 5 mō te x i te whārite \sqrt{40-3x}=x.

$$\sqrt{40-3\times 5}=5$$

Whakarūnātia. Ko te uara x=5 kua ngata te whārite.

$$5=5$$

Whakakapia te -8 mō te x i te whārite \sqrt{40-3x}=x.

$$\sqrt{40-3\left(-8\right)}=-8$$

Whakarūnātia. Ko te uara x=-8 kāore e ngata ana ki te whārite nā te mea e rerekē ngā tohu o te taha maui me te taha katau.

$$8=-8$$

Ko te whārite \sqrt{40-3x}=x he rongoā ahurei.

$$x=5$$