Whakaoti i te left(3x+2right)^(x+3)=x+4

$${ \left(3x+2 \right) }^{ } (x+3)=x+4$$

Whakaoti mō x

$$x=\frac{\sqrt{19}-5}{3}\approx -0.213700352$$

$$x=\frac{-\sqrt{19}-5}{3}\approx -3.119632981$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tātaihia te 3x+2 mā te pū o 1, kia riro ko 3x+2.

$$\left(3x+2\right)\left(x+3\right)=x+4$$

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

$$3x^{2}+11x+6=x+4$$

Tangohia te x mai i ngā taha e rua.

$$3x^{2}+11x+6-x=4$$

Pahekotia te 11x me -x, ka 10x.

$$3x^{2}+10x+6=4$$

Tangohia te 4 mai i ngā taha e rua.

$$3x^{2}+10x+6-4=0$$

Tangohia te 4 i te 6, ka 2.

$$3x^{2}+10x+2=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 3 mō a, 10 mō b, me 2 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\times 3\times 2}}{2\times 3}$$

Pūrua 10.

$$x=\frac{-10±\sqrt{100-4\times 3\times 2}}{2\times 3}$$

Whakareatia -4 ki te 3.

$$x=\frac{-10±\sqrt{100-12\times 2}}{2\times 3}$$

Whakareatia -12 ki te 2.

$$x=\frac{-10±\sqrt{100-24}}{2\times 3}$$

Tāpiri 100 ki te -24.

$$x=\frac{-10±\sqrt{76}}{2\times 3}$$

Tuhia te pūtakerua o te 76.

$$x=\frac{-10±2\sqrt{19}}{2\times 3}$$

Whakareatia 2 ki te 3.

$$x=\frac{-10±2\sqrt{19}}{6}$$

Nā, me whakaoti te whārite x=\frac{-10±2\sqrt{19}}{6} ina he tāpiri te ±. Tāpiri -10 ki te 2\sqrt{19}.

$$x=\frac{2\sqrt{19}-10}{6}$$

Whakawehe -10+2\sqrt{19} ki te 6.

$$x=\frac{\sqrt{19}-5}{3}$$

Nā, me whakaoti te whārite x=\frac{-10±2\sqrt{19}}{6} ina he tango te ±. Tango 2\sqrt{19} mai i -10.

$$x=\frac{-2\sqrt{19}-10}{6}$$

Whakawehe -10-2\sqrt{19} ki te 6.

$$x=\frac{-\sqrt{19}-5}{3}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{19}-5}{3} x=\frac{-\sqrt{19}-5}{3}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tātaihia te 3x+2 mā te pū o 1, kia riro ko 3x+2.

$$\left(3x+2\right)\left(x+3\right)=x+4$$

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

$$3x^{2}+11x+6=x+4$$

Tangohia te x mai i ngā taha e rua.

$$3x^{2}+11x+6-x=4$$

Pahekotia te 11x me -x, ka 10x.

$$3x^{2}+10x+6=4$$

Tangohia te 6 mai i ngā taha e rua.

$$3x^{2}+10x=4-6$$

Tangohia te 6 i te 4, ka -2.

$$3x^{2}+10x=-2$$

Whakawehea ngā taha e rua ki te 3.

$$\frac{3x^{2}+10x}{3}=-\frac{2}{3}$$

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

$$x^{2}+\frac{10}{3}x=-\frac{2}{3}$$

Whakawehea te \frac{10}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{5}{3}. Nā, tāpiria te pūrua o te \frac{5}{3} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{10}{3}x+\left(\frac{5}{3}\right)^{2}=-\frac{2}{3}+\left(\frac{5}{3}\right)^{2}$$

Pūruatia \frac{5}{3} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=-\frac{2}{3}+\frac{25}{9}$$

Tāpiri -\frac{2}{3} ki te \frac{25}{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.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=\frac{19}{9}$$

Tauwehea x^{2}+\frac{10}{3}x+\frac{25}{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}.

$$\left(x+\frac{5}{3}\right)^{2}=\frac{19}{9}$$

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

$$\sqrt{\left(x+\frac{5}{3}\right)^{2}}=\sqrt{\frac{19}{9}}$$

Whakarūnātia.

$$x+\frac{5}{3}=\frac{\sqrt{19}}{3} x+\frac{5}{3}=-\frac{\sqrt{19}}{3}$$

Me tango \frac{5}{3} mai i ngā taha e rua o te whārite.

$$x=\frac{\sqrt{19}-5}{3} x=\frac{-\sqrt{19}-5}{3}$$