Whakaoti i te f(x)=1/2x+3=x

$$f ( x ) = \frac { 1 } { 2 x + 3 } = x$$

Whakaoti mō x

$$x=\frac{\sqrt{17}-3}{4}\approx 0.280776406$$

$$x=\frac{-\sqrt{17}-3}{4}\approx -1.780776406$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tē taea kia ōrite te tāupe x ki -\frac{3}{2} nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te 2x+3.

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

Whakamahia te āhuatanga tohatoha hei whakarea te x ki te 2x+3.

$$1=2x^{2}+3x$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$2x^{2}+3x=1$$

Tangohia te 1 mai i ngā taha e rua.

$$2x^{2}+3x-1=0$$

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

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

Pūrua 3.

$$x=\frac{-3±\sqrt{9-4\times 2\left(-1\right)}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-3±\sqrt{9-8\left(-1\right)}}{2\times 2}$$

Whakareatia -8 ki te -1.

$$x=\frac{-3±\sqrt{9+8}}{2\times 2}$$

Tāpiri 9 ki te 8.

$$x=\frac{-3±\sqrt{17}}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{-3±\sqrt{17}}{4}$$

Nā, me whakaoti te whārite x=\frac{-3±\sqrt{17}}{4} ina he tāpiri te ±. Tāpiri -3 ki te \sqrt{17}.

$$x=\frac{\sqrt{17}-3}{4}$$

Nā, me whakaoti te whārite x=\frac{-3±\sqrt{17}}{4} ina he tango te ±. Tango \sqrt{17} mai i -3.

$$x=\frac{-\sqrt{17}-3}{4}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{17}-3}{4} x=\frac{-\sqrt{17}-3}{4}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tē taea kia ōrite te tāupe x ki -\frac{3}{2} nā te kore tautuhi i te whakawehenga mā te kore. Whakareatia ngā taha e rua o te whārite ki te 2x+3.

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

Whakamahia te āhuatanga tohatoha hei whakarea te x ki te 2x+3.

$$1=2x^{2}+3x$$

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$2x^{2}+3x=1$$

Whakawehea ngā taha e rua ki te 2.

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

Mā te whakawehe ki te 2 ka wetekia te whakareanga ki te 2.

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

Whakawehea te \frac{3}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{3}{4}. Nā, tāpiria te pūrua o te \frac{3}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia \frac{3}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{1}{2}+\frac{9}{16}$$

Tāpiri \frac{1}{2} ki te \frac{9}{16} 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.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{17}{16}$$

Tauwehea x^{2}+\frac{3}{2}x+\frac{9}{16}. 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}.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{17}{16}$$

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

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{17}{16}}$$

Whakarūnātia.

$$x+\frac{3}{4}=\frac{\sqrt{17}}{4} x+\frac{3}{4}=-\frac{\sqrt{17}}{4}$$

Me tango \frac{3}{4} mai i ngā taha e rua o te whārite.

$$x=\frac{\sqrt{17}-3}{4} x=\frac{-\sqrt{17}-3}{4}$$