Whakaoti i te -3x(2-x+4x)=1

$$- 3 x ( 2 - x + 4 x ) = 1$$

Whakaoti mō x

$$x=-\frac{1}{3}\approx -0.333333333$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Pahekotia te -x me 4x, ka 3x.

$$-3x\left(2+3x\right)=1$$

Whakamahia te āhuatanga tohatoha hei whakarea te -3x ki te 2+3x.

$$-6x-9x^{2}=1$$

Tangohia te 1 mai i ngā taha e rua.

$$-6x-9x^{2}-1=0$$

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.

$$-9x^{2}-6x-1=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -9 mō a, -6 mō b, me -1 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\left(-9\right)\left(-1\right)}}{2\left(-9\right)}$$

Pūrua -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\left(-9\right)\left(-1\right)}}{2\left(-9\right)}$$

Whakareatia -4 ki te -9.

$$x=\frac{-\left(-6\right)±\sqrt{36+36\left(-1\right)}}{2\left(-9\right)}$$

Whakareatia 36 ki te -1.

$$x=\frac{-\left(-6\right)±\sqrt{36-36}}{2\left(-9\right)}$$

Tāpiri 36 ki te -36.

$$x=\frac{-\left(-6\right)±\sqrt{0}}{2\left(-9\right)}$$

Tuhia te pūtakerua o te 0.

$$x=-\frac{-6}{2\left(-9\right)}$$

Ko te tauaro o -6 ko 6.

$$x=\frac{6}{2\left(-9\right)}$$

Whakareatia 2 ki te -9.

$$x=\frac{6}{-18}$$

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

$$x=-\frac{1}{3}$$

Ngā Upane mō te Whakaoti i te Pūrua

Pahekotia te -x me 4x, ka 3x.

$$-3x\left(2+3x\right)=1$$

Whakamahia te āhuatanga tohatoha hei whakarea te -3x ki te 2+3x.

$$-6x-9x^{2}=1$$

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.

$$-9x^{2}-6x=1$$

Whakawehea ngā taha e rua ki te -9.

$$\frac{-9x^{2}-6x}{-9}=\frac{1}{-9}$$

Mā te whakawehe ki te -9 ka wetekia te whakareanga ki te -9.

$$x^{2}+\left(-\frac{6}{-9}\right)x=\frac{1}{-9}$$

Whakahekea te hautanga \frac{-6}{-9} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 3.

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

Whakawehe 1 ki te -9.

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

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.

$$x^{2}+\frac{2}{3}x+\left(\frac{1}{3}\right)^{2}=-\frac{1}{9}+\left(\frac{1}{3}\right)^{2}$$

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

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

Tāpiri -\frac{1}{9} 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.

$$x^{2}+\frac{2}{3}x+\frac{1}{9}=0$$

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}.

$$\left(x+\frac{1}{3}\right)^{2}=0$$

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

$$\sqrt{\left(x+\frac{1}{3}\right)^{2}}=\sqrt{0}$$

Whakarūnātia.

$$x+\frac{1}{3}=0 x+\frac{1}{3}=0$$

Me tango \frac{1}{3} mai i ngā taha e rua o te whārite.

$$x=-\frac{1}{3} x=-\frac{1}{3}$$

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

$$x=-\frac{1}{3}$$