Whakaoti i te 12x(x+1)=-3

$$12 x ( x + 1 ) = - 3$$

Whakaoti mō x

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 12x ki te x+1.

$$12x^{2}+12x=-3$$

Me tāpiri te 3 ki ngā taha e rua.

$$12x^{2}+12x+3=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 12 mō a, 12 mō b, me 3 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\times 12\times 3}}{2\times 12}$$

Pūrua 12.

$$x=\frac{-12±\sqrt{144-4\times 12\times 3}}{2\times 12}$$

Whakareatia -4 ki te 12.

$$x=\frac{-12±\sqrt{144-48\times 3}}{2\times 12}$$

Whakareatia -48 ki te 3.

$$x=\frac{-12±\sqrt{144-144}}{2\times 12}$$

Tāpiri 144 ki te -144.

$$x=\frac{-12±\sqrt{0}}{2\times 12}$$

Tuhia te pūtakerua o te 0.

$$x=-\frac{12}{2\times 12}$$

Whakareatia 2 ki te 12.

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

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

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

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 12x ki te x+1.

$$12x^{2}+12x=-3$$

Whakawehea ngā taha e rua ki te 12.

$$\frac{12x^{2}+12x}{12}=-\frac{3}{12}$$

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

$$x^{2}+\frac{12}{12}x=-\frac{3}{12}$$

Whakawehe 12 ki te 12.

$$x^{2}+x=-\frac{3}{12}$$

Whakahekea te hautanga \frac{-3}{12} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 3.

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

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.

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

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

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

Tāpiri -\frac{1}{4} ki te \frac{1}{4} mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$x^{2}+x+\frac{1}{4}=0$$

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}.

$$\left(x+\frac{1}{2}\right)^{2}=0$$

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

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{0}$$

Whakarūnātia.

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

Me tango \frac{1}{2} mai i ngā taha e rua o te whārite.

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

Kua oti te whārite te whakatau. He ōrite ngā whakatau.

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