Whakaoti i te (3x-2)left(x+1right)=9

$$(3x-2) \left( x+1 \right) = 9$$

Whakaoti mō x

$$x = \frac{\sqrt{133} - 1}{6} \approx 1.755427099$$

$$x=\frac{-\sqrt{133}-1}{6}\approx -2.088760432$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tuaritanga hei whakarea te 3x-2 ki te x+1 ka whakakotahi i ngā kupu rite.

$$3x^{2}+x-2=9$$

Tangohia te 9 mai i ngā taha e rua.

$$3x^{2}+x-2-9=0$$

Tangohia te 9 i te -2, ka -11.

$$3x^{2}+x-11=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 3 mō a, 1 mō b, me -11 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times 3\left(-11\right)}}{2\times 3}$$

Pūrua 1.

$$x=\frac{-1±\sqrt{1-4\times 3\left(-11\right)}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-1±\sqrt{1-12\left(-11\right)}}{2\times 3}$$

Whakareatia -12 ki te -11.

$$x=\frac{-1±\sqrt{1+132}}{2\times 3}$$

Tāpiri 1 ki te 132.

$$x=\frac{-1±\sqrt{133}}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{-1±\sqrt{133}}{6}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{133}}{6} ina he tāpiri te ±. Tāpiri -1 ki te \sqrt{133}.

$$x=\frac{\sqrt{133}-1}{6}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{133}}{6} ina he tango te ±. Tango \sqrt{133} mai i -1.

$$x=\frac{-\sqrt{133}-1}{6}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{133}-1}{6} x=\frac{-\sqrt{133}-1}{6}$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tuaritanga hei whakarea te 3x-2 ki te x+1 ka whakakotahi i ngā kupu rite.

$$3x^{2}+x-2=9$$

Me tāpiri te 2 ki ngā taha e rua.

$$3x^{2}+x=9+2$$

Tāpirihia te 9 ki te 2, ka 11.

$$3x^{2}+x=11$$

Whakawehea ngā taha e rua ki te 3.

$$\frac{3x^{2}+x}{3}=\frac{11}{3}$$

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

$$x^{2}+\frac{1}{3}x=\frac{11}{3}$$

Whakawehea te \frac{1}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{6}. Nā, tāpiria te pūrua o te \frac{1}{6} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{1}{3}x+\left(\frac{1}{6}\right)^{2}=\frac{11}{3}+\left(\frac{1}{6}\right)^{2}$$

Pūruatia \frac{1}{6} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{11}{3}+\frac{1}{36}$$

Tāpiri \frac{11}{3} ki te \frac{1}{36} 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.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=\frac{133}{36}$$

Tauwehea x^{2}+\frac{1}{3}x+\frac{1}{36}. 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}.

$$\left(x+\frac{1}{6}\right)^{2}=\frac{133}{36}$$

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

$$\sqrt{\left(x+\frac{1}{6}\right)^{2}}=\sqrt{\frac{133}{36}}$$

Whakarūnātia.

$$x+\frac{1}{6}=\frac{\sqrt{133}}{6} x+\frac{1}{6}=-\frac{\sqrt{133}}{6}$$

Me tango \frac{1}{6} mai i ngā taha e rua o te whārite.

$$x=\frac{\sqrt{133}-1}{6} x=\frac{-\sqrt{133}-1}{6}$$