Whakaoti i te (5x-2)^2=5x^2-20x+4

$$( 5 x - 2 ) ^ { 2 } = 5 x ^ { 2 } - 20 x + 4$$

Whakaoti mō x

$$x=0$$

Ngā Upane mā te Kimi Pūtakerua

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(5x-2\right)^{2}.

$$25x^{2}-20x+4=5x^{2}-20x+4$$

Tangohia te 5x^{2} mai i ngā taha e rua.

$$25x^{2}-20x+4-5x^{2}=-20x+4$$

Pahekotia te 25x^{2} me -5x^{2}, ka 20x^{2}.

$$20x^{2}-20x+4=-20x+4$$

Me tāpiri te 20x ki ngā taha e rua.

$$20x^{2}-20x+4+20x=4$$

Pahekotia te -20x me 20x, ka 0.

$$20x^{2}+4=4$$

Tangohia te 4 mai i ngā taha e rua.

$$20x^{2}=4-4$$

Tangohia te 4 i te 4, ka 0.

$$20x^{2}=0$$

Whakawehea ngā taha e rua ki te 20. 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

Whakamahia te ture huarua \left(a-b\right)^{2}=a^{2}-2ab+b^{2} hei whakaroha \left(5x-2\right)^{2}.

$$25x^{2}-20x+4=5x^{2}-20x+4$$

Tangohia te 5x^{2} mai i ngā taha e rua.

$$25x^{2}-20x+4-5x^{2}=-20x+4$$

Pahekotia te 25x^{2} me -5x^{2}, ka 20x^{2}.

$$20x^{2}-20x+4=-20x+4$$

Me tāpiri te 20x ki ngā taha e rua.

$$20x^{2}-20x+4+20x=4$$

Pahekotia te -20x me 20x, ka 0.

$$20x^{2}+4=4$$

Tangohia te 4 mai i ngā taha e rua.

$$20x^{2}+4-4=0$$

Tangohia te 4 i te 4, ka 0.

$$20x^{2}=0$$

Whakawehea ngā taha e rua ki te 20. 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.

$$x=0$$