Whakaoti i te 4+x^2=4-x^2

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

Whakaoti mō x

$$x=0$$

Ngā Upane mā te Kimi Pūtakerua

Me tāpiri te x^{2} ki ngā taha e rua.

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

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

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

Tangohia te 4 mai i ngā taha e rua.

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

Tangohia te 4 i te 4, ka 0.

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

Whakawehea ngā taha e rua ki te 2. Ko te kore i whakawehea ki te tau ehara te kore ka hua ko te kore.

$$x^{2}=0$$

Tuhia te pūtakerua o ngā taha e rua o te whārite.

$$x=0 x=0$$

Kua oti te whārite te whakatau. He ōrite ngā whakatau.

$$x=0$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 4 mai i ngā taha e rua.

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

Tangohia te 4 i te 4, ka 0.

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

Me tāpiri te x^{2} ki ngā taha e rua.

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

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

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

Whakawehea ngā taha e rua ki te 2. Ko te kore i whakawehea ki te tau ehara te kore ka hua ko te kore.

$$x^{2}=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 0 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}}}{2}$$

Tuhia te pūtakerua o te 0^{2}.

$$x=\frac{0±0}{2}$$

Whakawehe 0 ki te 2.

$$x=0$$