Whakaoti i te 6x^2=8x

$$6 x ^ { 2 } = 8 x$$

Whakaoti mō x

$$x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Tangohia te 8x mai i ngā taha e rua.

$$6x^{2}-8x=0$$

Tauwehea te x.

$$x\left(6x-8\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te 6x-8=0.

$$x=0 x=\frac{4}{3}$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 8x mai i ngā taha e rua.

$$6x^{2}-8x=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 6 mō a, -8 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}}}{2\times 6}$$

Tuhia te pūtakerua o te \left(-8\right)^{2}.

$$x=\frac{-\left(-8\right)±8}{2\times 6}$$

Ko te tauaro o -8 ko 8.

$$x=\frac{8±8}{2\times 6}$$

Whakareatia 2 ki te 6.

$$x=\frac{8±8}{12}$$

Nā, me whakaoti te whārite x=\frac{8±8}{12} ina he tāpiri te ±. Tāpiri 8 ki te 8.

$$x=\frac{16}{12}$$

Whakahekea te hautanga \frac{16}{12} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$x=\frac{4}{3}$$

Nā, me whakaoti te whārite x=\frac{8±8}{12} ina he tango te ±. Tango 8 mai i 8.

$$x=\frac{0}{12}$$

Whakawehe 0 ki te 12.

$$x=0$$

Kua oti te whārite te whakatau.

$$x=\frac{4}{3} x=0$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 8x mai i ngā taha e rua.

$$6x^{2}-8x=0$$

Whakawehea ngā taha e rua ki te 6.

$$\frac{6x^{2}-8x}{6}=\frac{0}{6}$$

Mā te whakawehe ki te 6 ka wetekia te whakareanga ki te 6.

$$x^{2}+\left(-\frac{8}{6}\right)x=\frac{0}{6}$$

Whakahekea te hautanga \frac{-8}{6} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x^{2}-\frac{4}{3}x=\frac{0}{6}$$

Whakawehe 0 ki te 6.

$$x^{2}-\frac{4}{3}x=0$$

Whakawehea te -\frac{4}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{2}{3}. Nā, tāpiria te pūrua o te -\frac{2}{3} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia -\frac{2}{3} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{4}{9}$$

Tauwehea x^{2}-\frac{4}{3}x+\frac{4}{9}. 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}.

$$\left(x-\frac{2}{3}\right)^{2}=\frac{4}{9}$$

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

$$\sqrt{\left(x-\frac{2}{3}\right)^{2}}=\sqrt{\frac{4}{9}}$$

Whakarūnātia.

$$x-\frac{2}{3}=\frac{2}{3} x-\frac{2}{3}=-\frac{2}{3}$$

Me tāpiri \frac{2}{3} ki ngā taha e rua o te whārite.

$$x=\frac{4}{3} x=0$$