Whakaoti i te 6x(x-1)=0

$$6 x ( x - 1 ) = 0$$

Whakaoti mō x

$$x=1$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Whakamahia te āhuatanga tohatoha hei whakarea te 6x ki te x-1.

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

Tauwehea te x.

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

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

$$x=0 x=1$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 6x ki te x-1.

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

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

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

Tuhia te pūtakerua o te \left(-6\right)^{2}.

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

Ko te tauaro o -6 ko 6.

$$x=\frac{6±6}{2\times 6}$$

Whakareatia 2 ki te 6.

$$x=\frac{6±6}{12}$$

Nā, me whakaoti te whārite x=\frac{6±6}{12} ina he tāpiri te ±. Tāpiri 6 ki te 6.

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

Whakawehe 12 ki te 12.

$$x=1$$

Nā, me whakaoti te whārite x=\frac{6±6}{12} ina he tango te ±. Tango 6 mai i 6.

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

Whakawehe 0 ki te 12.

$$x=0$$

Kua oti te whārite te whakatau.

$$x=1 x=0$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 6x ki te x-1.

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

Whakawehea ngā taha e rua ki te 6.

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

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

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

Whakawehe -6 ki te 6.

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

Whakawehe 0 ki te 6.

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

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

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

Pūruatia -\frac{1}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-x+\frac{1}{4}=\frac{1}{4}$$

Tauwehea x^{2}-x+\frac{1}{4}. 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}.

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

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

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

Whakarūnātia.

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

Me tāpiri \frac{1}{2} ki ngā taha e rua o te whārite.

$$x=1 x=0$$