Whakaoti i te x+2=sqrt{4-x^2}

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

Whakaoti mō x

$$x=-2$$

$$x=0$$

Ngā Upane Otinga

Pūruatia ngā taha e rua o te whārite.

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

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

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

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

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

Tangohia te 4 mai i ngā taha e rua.

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

Tangohia te 4 i te 4, ka 0.

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

Me tāpiri te x^{2} ki ngā taha e rua.

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

Pahekotia te x^{2} me x^{2}, ka 2x^{2}.

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

Tauwehea te x.

$$x\left(2x+4\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te 2x+4=0.

$$x=0 x=-2$$

Whakakapia te 0 mō te x i te whārite x+2=\sqrt{4-x^{2}}.

$$0+2=\sqrt{4-0^{2}}$$

Whakarūnātia. Ko te uara x=0 kua ngata te whārite.

$$2=2$$

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

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

Whakarūnātia. Ko te uara x=-2 kua ngata te whārite.

$$0=0$$

Rārangihia ngā rongoā katoa o x+2=\sqrt{4-x^{2}}.

$$x=0 x=-2$$