Whakaoti i te frac{-2sqrt{x-4}}{-2}=x-4/-2

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

Whakaoti mō x

$$x=4$$

Ngā Upane Otinga

Whakareatia ngā taha e rua o te whārite ki te -2.

$$-2\sqrt{x-4}=x-4$$

Tangohia te x mai i ngā taha e rua.

$$-2\sqrt{x-4}-x=-4$$

Me tango -x mai i ngā taha e rua o te whārite.

$$-2\sqrt{x-4}=-4+x$$

Pūruatia ngā taha e rua o te whārite.

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

Whakarohaina te \left(-2\sqrt{x-4}\right)^{2}.

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

Tātaihia te -2 mā te pū o 2, kia riro ko 4.

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

Tātaihia te \sqrt{x-4} mā te pū o 2, kia riro ko x-4.

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

Whakamahia te āhuatanga tohatoha hei whakarea te 4 ki te x-4.

$$4x-16=\left(-4+x\right)^{2}$$

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(-4+x\right)^{2}.

$$4x-16=16-8x+x^{2}$$

Me tāpiri te 8x ki ngā taha e rua.

$$4x-16+8x=16+x^{2}$$

Pahekotia te 4x me 8x, ka 12x.

$$12x-16=16+x^{2}$$

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

$$12x-16-x^{2}=16$$

Tangohia te 16 mai i ngā taha e rua.

$$12x-16-x^{2}-16=0$$

Tangohia te 16 i te -16, ka -32.

$$12x-32-x^{2}=0$$

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

$$-x^{2}+12x-32=0$$

Hei whakaoti i te whārite, whakatauwehea te taha mauī mā te whakarōpū. Tuatahi, me tuhi anō te taha mauī hei -x^{2}+ax+bx-32. Hei kimi a me b, whakaritea tētahi pūnaha kia whakaoti.

$$a+b=12 ab=-\left(-32\right)=32$$

I te mea kua tōrunga te ab, he ōrite te tohu o a me b. I te mea kua tōrunga te a+b, he tōrunga hoki a a me b. Whakarārangitia ngā tau tōpū takirua pērā katoa ka hoatu i te hua 32.

$$1,32 2,16 4,8$$

Tātaihia te tapeke mō ia takirua.

$$1+32=33 2+16=18 4+8=12$$

Ko te otinga te takirua ka hoatu i te tapeke 12.

$$a=8 b=4$$

Tuhia anō te -x^{2}+12x-32 hei \left(-x^{2}+8x\right)+\left(4x-32\right).

$$\left(-x^{2}+8x\right)+\left(4x-32\right)$$

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

$$-x\left(x-8\right)+4\left(x-8\right)$$

Whakatauwehea atu te kīanga pātahi x-8 mā te whakamahi i te āhuatanga tātai tohatoha.

$$\left(x-8\right)\left(-x+4\right)$$

Hei kimi otinga whārite, me whakaoti te x-8=0 me te -x+4=0.

$$x=8 x=4$$

Whakakapia te 8 mō te x i te whārite \frac{-2\sqrt{x-4}}{-2}=\frac{x-4}{-2}.

$$\frac{-2\sqrt{8-4}}{-2}=\frac{8-4}{-2}$$

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.

$$2=-2$$

Whakakapia te 4 mō te x i te whārite \frac{-2\sqrt{x-4}}{-2}=\frac{x-4}{-2}.

$$\frac{-2\sqrt{4-4}}{-2}=\frac{4-4}{-2}$$

Whakarūnātia. Ko te uara x=4 kua ngata te whārite.

$$0=0$$

Ko te whārite -2\sqrt{x-4}=x-4 he rongoā ahurei.

$$x=4$$