Whakaoti i te (3x+2)(x+2)-4(x+1)^2=0

$$( 3 x + 2 ) ( x + 2 ) - 4 ( x + 1 ) ^ { 2 } = 0$$

Whakaoti mō x

$$x=0$$

Ngā Upane mā te Kimi Pūtakerua

Whakamahia te āhuatanga tuaritanga hei whakarea te 3x+2 ki te x+2 ka whakakotahi i ngā kupu rite.

$$3x^{2}+8x+4-4\left(x+1\right)^{2}=0$$

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(x+1\right)^{2}.

$$3x^{2}+8x+4-4\left(x^{2}+2x+1\right)=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te -4 ki te x^{2}+2x+1.

$$3x^{2}+8x+4-4x^{2}-8x-4=0$$

Pahekotia te 3x^{2} me -4x^{2}, ka -x^{2}.

$$-x^{2}+8x+4-8x-4=0$$

Pahekotia te 8x me -8x, ka 0.

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

Tangohia te 4 i te 4, ka 0.

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

Whakawehea ngā taha e rua ki te -1. 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 āhuatanga tuaritanga hei whakarea te 3x+2 ki te x+2 ka whakakotahi i ngā kupu rite.

$$3x^{2}+8x+4-4\left(x+1\right)^{2}=0$$

Whakamahia te ture huarua \left(a+b\right)^{2}=a^{2}+2ab+b^{2} hei whakaroha \left(x+1\right)^{2}.

$$3x^{2}+8x+4-4\left(x^{2}+2x+1\right)=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te -4 ki te x^{2}+2x+1.

$$3x^{2}+8x+4-4x^{2}-8x-4=0$$

Pahekotia te 3x^{2} me -4x^{2}, ka -x^{2}.

$$-x^{2}+8x+4-8x-4=0$$

Pahekotia te 8x me -8x, ka 0.

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

Tangohia te 4 i te 4, ka 0.

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

Whakawehea ngā taha e rua ki te -1. 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.

$$x=0$$