Whakaoti i te 6x(2x+1)=0

$$6 x ( 2 x + 1 ) = 0$$

Whakaoti mō x

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

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Whakamahia te āhuatanga tohatoha hei whakarea te 6x ki te 2x+1.

$$12x^{2}+6x=0$$

Tauwehea te x.

$$x\left(12x+6\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te 12x+6=0.

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 6x ki te 2x+1.

$$12x^{2}+6x=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 12 mō a, 6 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}}}{2\times 12}$$

Tuhia te pūtakerua o te 6^{2}.

$$x=\frac{-6±6}{2\times 12}$$

Whakareatia 2 ki te 12.

$$x=\frac{-6±6}{24}$$

Nā, me whakaoti te whārite x=\frac{-6±6}{24} ina he tāpiri te ±. Tāpiri -6 ki te 6.

$$x=\frac{0}{24}$$

Whakawehe 0 ki te 24.

$$x=0$$

Nā, me whakaoti te whārite x=\frac{-6±6}{24} ina he tango te ±. Tango 6 mai i -6.

$$x=-\frac{12}{24}$$

Whakahekea te hautanga \frac{-12}{24} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 12.

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

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 6x ki te 2x+1.

$$12x^{2}+6x=0$$

Whakawehea ngā taha e rua ki te 12.

$$\frac{12x^{2}+6x}{12}=\frac{0}{12}$$

Mā te whakawehe ki te 12 ka wetekia te whakareanga ki te 12.

$$x^{2}+\frac{6}{12}x=\frac{0}{12}$$

Whakahekea te hautanga \frac{6}{12} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

$$x^{2}+\frac{1}{2}x=\frac{0}{12}$$

Whakawehe 0 ki te 12.

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

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 tango \frac{1}{4} mai i ngā taha e rua o te whārite.

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