Whakaoti i te x^2+4x-5=4x-5

$$x ^ { 2 } + 4 x - 5 = 4 x - 5$$

Whakaoti mō x

$$x=0$$

Ngā Upane mā te Kimi Pūtakerua

Tangohia te 4x mai i ngā taha e rua.

$$x^{2}+4x-5-4x=-5$$

Pahekotia te 4x me -4x, ka 0.

$$x^{2}-5=-5$$

Me tāpiri te 5 ki ngā taha e rua.

$$x^{2}=-5+5$$

Tāpirihia te -5 ki te 5, ka 0.

$$x^{2}=0$$

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 4x mai i ngā taha e rua.

$$x^{2}+4x-5-4x=-5$$

Pahekotia te 4x me -4x, ka 0.

$$x^{2}-5=-5$$

Me tāpiri te 5 ki ngā taha e rua.

$$x^{2}-5+5=0$$

Tāpirihia te -5 ki te 5, ka 0.

$$x^{2}=0$$

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.

$$x=0$$