Whakaoti i te (4x+3)(2-3x)=6

$$(4x+3)(2-3x)=6$$

Whakaoti mō x

$$x=-\frac{1}{12}\approx -0.083333333$$

$$x=0$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tuaritanga hei whakarea te 4x+3 ki te 2-3x ka whakakotahi i ngā kupu rite.

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

Tangohia te 6 mai i ngā taha e rua.

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

Tangohia te 6 i te 6, ka 0.

$$-x-12x^{2}=0$$

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.

$$-12x^{2}-x=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -12 mō a, -1 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1}}{2\left(-12\right)}$$

Tuhia te pūtakerua o te 1.

$$x=\frac{-\left(-1\right)±1}{2\left(-12\right)}$$

Ko te tauaro o -1 ko 1.

$$x=\frac{1±1}{2\left(-12\right)}$$

Whakareatia 2 ki te -12.

$$x=\frac{1±1}{-24}$$

Nā, me whakaoti te whārite x=\frac{1±1}{-24} ina he tāpiri te ±. Tāpiri 1 ki te 1.

$$x=\frac{2}{-24}$$

Whakahekea te hautanga \frac{2}{-24} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x=-\frac{1}{12}$$

Nā, me whakaoti te whārite x=\frac{1±1}{-24} ina he tango te ±. Tango 1 mai i 1.

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

Whakawehe 0 ki te -24.

$$x=0$$

Kua oti te whārite te whakatau.

$$x=-\frac{1}{12} x=0$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tuaritanga hei whakarea te 4x+3 ki te 2-3x ka whakakotahi i ngā kupu rite.

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

Tangohia te 6 mai i ngā taha e rua.

$$-x-12x^{2}=6-6$$

Tangohia te 6 i te 6, ka 0.

$$-x-12x^{2}=0$$

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.

$$-12x^{2}-x=0$$

Whakawehea ngā taha e rua ki te -12.

$$\frac{-12x^{2}-x}{-12}=\frac{0}{-12}$$

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

$$x^{2}+\left(-\frac{1}{-12}\right)x=\frac{0}{-12}$$

Whakawehe -1 ki te -12.

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

Whakawehe 0 ki te -12.

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

Whakawehea te \frac{1}{12}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{24}. Nā, tāpiria te pūrua o te \frac{1}{24} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia \frac{1}{24} mā te pūrua i te taurunga me te tauraro o te hautanga.

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

Tauwehea x^{2}+\frac{1}{12}x+\frac{1}{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}.

$$\left(x+\frac{1}{24}\right)^{2}=\frac{1}{576}$$

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

$$\sqrt{\left(x+\frac{1}{24}\right)^{2}}=\sqrt{\frac{1}{576}}$$

Whakarūnātia.

$$x+\frac{1}{24}=\frac{1}{24} x+\frac{1}{24}=-\frac{1}{24}$$

Me tango \frac{1}{24} mai i ngā taha e rua o te whārite.

$$x=0 x=-\frac{1}{12}$$