Whakaoti i te 3x(x-1)+4x=3/4(x+1)-6x

$$3 x ( x - 1 ) + 4 x = \frac { 3 } { 4 } ( x + 1 ) - 6 x$$

Whakaoti mō x

$$x=\frac{\sqrt{769}-25}{24}\approx 0.113785385$$

$$x=\frac{-\sqrt{769}-25}{24}\approx -2.197118719$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 3x ki te x-1.

$$3x^{2}-3x+4x=\frac{3}{4}\left(x+1\right)-6x$$

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

$$3x^{2}+x=\frac{3}{4}\left(x+1\right)-6x$$

Whakamahia te āhuatanga tohatoha hei whakarea te \frac{3}{4} ki te x+1.

$$3x^{2}+x=\frac{3}{4}x+\frac{3}{4}-6x$$

Pahekotia te \frac{3}{4}x me -6x, ka -\frac{21}{4}x.

$$3x^{2}+x=-\frac{21}{4}x+\frac{3}{4}$$

Me tāpiri te \frac{21}{4}x ki ngā taha e rua.

$$3x^{2}+x+\frac{21}{4}x=\frac{3}{4}$$

Pahekotia te x me \frac{21}{4}x, ka \frac{25}{4}x.

$$3x^{2}+\frac{25}{4}x=\frac{3}{4}$$

Tangohia te \frac{3}{4} mai i ngā taha e rua.

$$3x^{2}+\frac{25}{4}x-\frac{3}{4}=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 3 mō a, \frac{25}{4} mō b, me -\frac{3}{4} mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{25}{4}±\sqrt{\left(\frac{25}{4}\right)^{2}-4\times 3\left(-\frac{3}{4}\right)}}{2\times 3}$$

Pūruatia \frac{25}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x=\frac{-\frac{25}{4}±\sqrt{\frac{625}{16}-4\times 3\left(-\frac{3}{4}\right)}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-\frac{25}{4}±\sqrt{\frac{625}{16}-12\left(-\frac{3}{4}\right)}}{2\times 3}$$

Whakareatia -12 ki te -\frac{3}{4}.

$$x=\frac{-\frac{25}{4}±\sqrt{\frac{625}{16}+9}}{2\times 3}$$

Tāpiri \frac{625}{16} ki te 9.

$$x=\frac{-\frac{25}{4}±\sqrt{\frac{769}{16}}}{2\times 3}$$

Tuhia te pūtakerua o te \frac{769}{16}.

$$x=\frac{-\frac{25}{4}±\frac{\sqrt{769}}{4}}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{-\frac{25}{4}±\frac{\sqrt{769}}{4}}{6}$$

Nā, me whakaoti te whārite x=\frac{-\frac{25}{4}±\frac{\sqrt{769}}{4}}{6} ina he tāpiri te ±. Tāpiri -\frac{25}{4} ki te \frac{\sqrt{769}}{4}.

$$x=\frac{\sqrt{769}-25}{4\times 6}$$

Whakawehe \frac{-25+\sqrt{769}}{4} ki te 6.

$$x=\frac{\sqrt{769}-25}{24}$$

Nā, me whakaoti te whārite x=\frac{-\frac{25}{4}±\frac{\sqrt{769}}{4}}{6} ina he tango te ±. Tango \frac{\sqrt{769}}{4} mai i -\frac{25}{4}.

$$x=\frac{-\sqrt{769}-25}{4\times 6}$$

Whakawehe \frac{-25-\sqrt{769}}{4} ki te 6.

$$x=\frac{-\sqrt{769}-25}{24}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{769}-25}{24} x=\frac{-\sqrt{769}-25}{24}$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 3x ki te x-1.

$$3x^{2}-3x+4x=\frac{3}{4}\left(x+1\right)-6x$$

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

$$3x^{2}+x=\frac{3}{4}\left(x+1\right)-6x$$

Whakamahia te āhuatanga tohatoha hei whakarea te \frac{3}{4} ki te x+1.

$$3x^{2}+x=\frac{3}{4}x+\frac{3}{4}-6x$$

Pahekotia te \frac{3}{4}x me -6x, ka -\frac{21}{4}x.

$$3x^{2}+x=-\frac{21}{4}x+\frac{3}{4}$$

Me tāpiri te \frac{21}{4}x ki ngā taha e rua.

$$3x^{2}+x+\frac{21}{4}x=\frac{3}{4}$$

Pahekotia te x me \frac{21}{4}x, ka \frac{25}{4}x.

$$3x^{2}+\frac{25}{4}x=\frac{3}{4}$$

Whakawehea ngā taha e rua ki te 3.

$$\frac{3x^{2}+\frac{25}{4}x}{3}=\frac{\frac{3}{4}}{3}$$

Mā te whakawehe ki te 3 ka wetekia te whakareanga ki te 3.

$$x^{2}+\frac{\frac{25}{4}}{3}x=\frac{\frac{3}{4}}{3}$$

Whakawehe \frac{25}{4} ki te 3.

$$x^{2}+\frac{25}{12}x=\frac{\frac{3}{4}}{3}$$

Whakawehe \frac{3}{4} ki te 3.

$$x^{2}+\frac{25}{12}x=\frac{1}{4}$$

Whakawehea te \frac{25}{12}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{25}{24}. Nā, tāpiria te pūrua o te \frac{25}{24} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{25}{12}x+\left(\frac{25}{24}\right)^{2}=\frac{1}{4}+\left(\frac{25}{24}\right)^{2}$$

Pūruatia \frac{25}{24} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{25}{12}x+\frac{625}{576}=\frac{1}{4}+\frac{625}{576}$$

Tāpiri \frac{1}{4} ki te \frac{625}{576} 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.

$$x^{2}+\frac{25}{12}x+\frac{625}{576}=\frac{769}{576}$$

Tauwehea x^{2}+\frac{25}{12}x+\frac{625}{576}. 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}.

$$\left(x+\frac{25}{24}\right)^{2}=\frac{769}{576}$$

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

$$\sqrt{\left(x+\frac{25}{24}\right)^{2}}=\sqrt{\frac{769}{576}}$$

Whakarūnātia.

$$x+\frac{25}{24}=\frac{\sqrt{769}}{24} x+\frac{25}{24}=-\frac{\sqrt{769}}{24}$$

Me tango \frac{25}{24} mai i ngā taha e rua o te whārite.

$$x=\frac{\sqrt{769}-25}{24} x=\frac{-\sqrt{769}-25}{24}$$