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

$$3 ( x + 1 ) ( x - 2 ) = 2 ( x ^ { 2 } - 2 )$$

Whakaoti mō x

$$x = \frac{\sqrt{17} + 3}{2} \approx 3.561552813$$

$$x=\frac{3-\sqrt{17}}{2}\approx -0.561552813$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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

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

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

$$3x^{2}-3x-6=2\left(x^{2}-2\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te x^{2}-2.

$$3x^{2}-3x-6=2x^{2}-4$$

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

$$3x^{2}-3x-6-2x^{2}=-4$$

Pahekotia te 3x^{2} me -2x^{2}, ka x^{2}.

$$x^{2}-3x-6=-4$$

Me tāpiri te 4 ki ngā taha e rua.

$$x^{2}-3x-6+4=0$$

Tāpirihia te -6 ki te 4, ka -2.

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

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

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

Pūrua -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-2\right)}}{2}$$

Whakareatia -4 ki te -2.

$$x=\frac{-\left(-3\right)±\sqrt{9+8}}{2}$$

Tāpiri 9 ki te 8.

$$x=\frac{-\left(-3\right)±\sqrt{17}}{2}$$

Ko te tauaro o -3 ko 3.

$$x=\frac{3±\sqrt{17}}{2}$$

Nā, me whakaoti te whārite x=\frac{3±\sqrt{17}}{2} ina he tāpiri te ±. Tāpiri 3 ki te \sqrt{17}.

$$x=\frac{\sqrt{17}+3}{2}$$

Nā, me whakaoti te whārite x=\frac{3±\sqrt{17}}{2} ina he tango te ±. Tango \sqrt{17} mai i 3.

$$x=\frac{3-\sqrt{17}}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{17}+3}{2} x=\frac{3-\sqrt{17}}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

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

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

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

$$3x^{2}-3x-6=2\left(x^{2}-2\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 2 ki te x^{2}-2.

$$3x^{2}-3x-6=2x^{2}-4$$

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

$$3x^{2}-3x-6-2x^{2}=-4$$

Pahekotia te 3x^{2} me -2x^{2}, ka x^{2}.

$$x^{2}-3x-6=-4$$

Me tāpiri te 6 ki ngā taha e rua.

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

Tāpirihia te -4 ki te 6, ka 2.

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

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

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

Pūruatia -\frac{3}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-3x+\frac{9}{4}=2+\frac{9}{4}$$

Tāpiri 2 ki te \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{17}{4}$$

Tauwehea x^{2}-3x+\frac{9}{4}. 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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{17}{4}$$

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

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{17}{4}}$$

Whakarūnātia.

$$x-\frac{3}{2}=\frac{\sqrt{17}}{2} x-\frac{3}{2}=-\frac{\sqrt{17}}{2}$$

Me tāpiri \frac{3}{2} ki ngā taha e rua o te whārite.

$$x=\frac{\sqrt{17}+3}{2} x=\frac{3-\sqrt{17}}{2}$$