Whakaoti i te x^2-3x=3x^2-4x

$${ x }^{ 2 } -3x = 3 { x }^{ 2 } -4x$$

Whakaoti mō x

$$x=\frac{1}{2}=0.5$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Tangohia te 3x^{2} mai i ngā taha e rua.

$$x^{2}-3x-3x^{2}=-4x$$

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

$$-2x^{2}-3x=-4x$$

Me tāpiri te 4x ki ngā taha e rua.

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

Pahekotia te -3x me 4x, ka x.

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

Tauwehea te x.

$$x\left(-2x+1\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te -2x+1=0.

$$x=0 x=\frac{1}{2}$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 3x^{2} mai i ngā taha e rua.

$$x^{2}-3x-3x^{2}=-4x$$

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

$$-2x^{2}-3x=-4x$$

Me tāpiri te 4x ki ngā taha e rua.

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

Pahekotia te -3x me 4x, ka x.

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

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

$$x=\frac{-1±\sqrt{1^{2}}}{2\left(-2\right)}$$

Tuhia te pūtakerua o te 1^{2}.

$$x=\frac{-1±1}{2\left(-2\right)}$$

Whakareatia 2 ki te -2.

$$x=\frac{-1±1}{-4}$$

Nā, me whakaoti te whārite x=\frac{-1±1}{-4} ina he tāpiri te ±. Tāpiri -1 ki te 1.

$$x=\frac{0}{-4}$$

Whakawehe 0 ki te -4.

$$x=0$$

Nā, me whakaoti te whārite x=\frac{-1±1}{-4} ina he tango te ±. Tango 1 mai i -1.

$$x=-\frac{2}{-4}$$

Whakahekea te hautanga \frac{-2}{-4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=\frac{1}{2}$$

Kua oti te whārite te whakatau.

$$x=0 x=\frac{1}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 3x^{2} mai i ngā taha e rua.

$$x^{2}-3x-3x^{2}=-4x$$

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

$$-2x^{2}-3x=-4x$$

Me tāpiri te 4x ki ngā taha e rua.

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

Pahekotia te -3x me 4x, ka x.

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

Whakawehea ngā taha e rua ki te -2.

$$\frac{-2x^{2}+x}{-2}=\frac{0}{-2}$$

Mā te whakawehe ki te -2 ka wetekia te whakareanga ki te -2.

$$x^{2}+\frac{1}{-2}x=\frac{0}{-2}$$

Whakawehe 1 ki te -2.

$$x^{2}-\frac{1}{2}x=\frac{0}{-2}$$

Whakawehe 0 ki te -2.

$$x^{2}-\frac{1}{2}x=0$$

Whakawehea te -\frac{1}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{4}. Nā, tāpiria te pūrua o te -\frac{1}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia -\frac{1}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{1}{16}$$

Tauwehea x^{2}-\frac{1}{2}x+\frac{1}{16}. Ko te tikanga pūnoa, ina ko x^{2}+bx+c he pūrua tika pūrua tika pū, ka taea taua mea te tauwehea i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{1}{16}$$

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

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Whakarūnātia.

$$x-\frac{1}{4}=\frac{1}{4} x-\frac{1}{4}=-\frac{1}{4}$$

Me tāpiri \frac{1}{4} ki ngā taha e rua o te whārite.

$$x=\frac{1}{2} x=0$$