Whakaoti i te 4x+6/12x+4=2x/6

$$\frac { 4 x + 6 } { 12 x + 4 } = \frac { 2 x } { 6 }$$

Whakaoti mō x

$$x=\frac{\sqrt{58}}{6}+\frac{1}{3}\approx 1.602628851$$

$$x=-\frac{\sqrt{58}}{6}+\frac{1}{3}\approx -0.935962184$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki -\frac{1}{3} nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te 12\left(3x+1\right), arā, te tauraro pātahi he tino iti rawa te kitea o 12x+4,6.

$$3\left(4x+6\right)=\left(6x+2\right)\times 2x$$

Whakamahia te āhuatanga tohatoha hei whakarea te 3 ki te 4x+6.

$$12x+18=\left(6x+2\right)\times 2x$$

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

$$12x+18=\left(12x+4\right)x$$

Whakamahia te āhuatanga tohatoha hei whakarea te 12x+4 ki te x.

$$12x+18=12x^{2}+4x$$

Tangohia te 12x^{2} mai i ngā taha e rua.

$$12x+18-12x^{2}=4x$$

Tangohia te 4x mai i ngā taha e rua.

$$12x+18-12x^{2}-4x=0$$

Pahekotia te 12x me -4x, ka 8x.

$$8x+18-12x^{2}=0$$

Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.

$$-12x^{2}+8x+18=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -12 mō a, 8 mō b, me 18 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\left(-12\right)\times 18}}{2\left(-12\right)}$$

Pūrua 8.

$$x=\frac{-8±\sqrt{64-4\left(-12\right)\times 18}}{2\left(-12\right)}$$

Whakareatia -4 ki te -12.

$$x=\frac{-8±\sqrt{64+48\times 18}}{2\left(-12\right)}$$

Whakareatia 48 ki te 18.

$$x=\frac{-8±\sqrt{64+864}}{2\left(-12\right)}$$

Tāpiri 64 ki te 864.

$$x=\frac{-8±\sqrt{928}}{2\left(-12\right)}$$

Tuhia te pūtakerua o te 928.

$$x=\frac{-8±4\sqrt{58}}{2\left(-12\right)}$$

Whakareatia 2 ki te -12.

$$x=\frac{-8±4\sqrt{58}}{-24}$$

Nā, me whakaoti te whārite x=\frac{-8±4\sqrt{58}}{-24} ina he tāpiri te ±. Tāpiri -8 ki te 4\sqrt{58}.

$$x=\frac{4\sqrt{58}-8}{-24}$$

Whakawehe -8+4\sqrt{58} ki te -24.

$$x=-\frac{\sqrt{58}}{6}+\frac{1}{3}$$

Nā, me whakaoti te whārite x=\frac{-8±4\sqrt{58}}{-24} ina he tango te ±. Tango 4\sqrt{58} mai i -8.

$$x=\frac{-4\sqrt{58}-8}{-24}$$

Whakawehe -8-4\sqrt{58} ki te -24.

$$x=\frac{\sqrt{58}}{6}+\frac{1}{3}$$

Kua oti te whārite te whakatau.

$$x=-\frac{\sqrt{58}}{6}+\frac{1}{3} x=\frac{\sqrt{58}}{6}+\frac{1}{3}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki -\frac{1}{3} nā te kore tautuhi i te whakawehenga mā te kore. Me whakarea ngā taha e rua o te whārite ki te 12\left(3x+1\right), arā, te tauraro pātahi he tino iti rawa te kitea o 12x+4,6.

$$3\left(4x+6\right)=\left(6x+2\right)\times 2x$$

Whakamahia te āhuatanga tohatoha hei whakarea te 3 ki te 4x+6.

$$12x+18=\left(6x+2\right)\times 2x$$

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

$$12x+18=\left(12x+4\right)x$$

Whakamahia te āhuatanga tohatoha hei whakarea te 12x+4 ki te x.

$$12x+18=12x^{2}+4x$$

Tangohia te 12x^{2} mai i ngā taha e rua.

$$12x+18-12x^{2}=4x$$

Tangohia te 4x mai i ngā taha e rua.

$$12x+18-12x^{2}-4x=0$$

Pahekotia te 12x me -4x, ka 8x.

$$8x+18-12x^{2}=0$$

Tangohia te 18 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$8x-12x^{2}=-18$$

Ko ngā whārite pūrua pēnei i tēnei nā ka taea te whakaoti mā te whakaoti i te pūrua. Hei whakaoti i te pūrua, ko te whārite me mātua tuhi ki te āhua x^{2}+bx=c.

$$-12x^{2}+8x=-18$$

Whakawehea ngā taha e rua ki te -12.

$$\frac{-12x^{2}+8x}{-12}=-\frac{18}{-12}$$

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

$$x^{2}+\frac{8}{-12}x=-\frac{18}{-12}$$

Whakahekea te hautanga \frac{8}{-12} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$x^{2}-\frac{2}{3}x=-\frac{18}{-12}$$

Whakahekea te hautanga \frac{-18}{-12} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

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

Whakawehea te -\frac{2}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{3}. Nā, tāpiria te pūrua o te -\frac{1}{3} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{2}{3}x+\left(-\frac{1}{3}\right)^{2}=\frac{3}{2}+\left(-\frac{1}{3}\right)^{2}$$

Pūruatia -\frac{1}{3} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{3}{2}+\frac{1}{9}$$

Tāpiri \frac{3}{2} ki te \frac{1}{9} 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.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{29}{18}$$

Tauwehea x^{2}-\frac{2}{3}x+\frac{1}{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}.

$$\left(x-\frac{1}{3}\right)^{2}=\frac{29}{18}$$

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

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{\frac{29}{18}}$$

Whakarūnātia.

$$x-\frac{1}{3}=\frac{\sqrt{58}}{6} x-\frac{1}{3}=-\frac{\sqrt{58}}{6}$$

Me tāpiri \frac{1}{3} ki ngā taha e rua o te whārite.

$$x=\frac{\sqrt{58}}{6}+\frac{1}{3} x=-\frac{\sqrt{58}}{6}+\frac{1}{3}$$