Whakaoti i te -x+3/4=-x^2+2x+3

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

Whakaoti mō x

$$x = \frac{3 \sqrt{2} + 3}{2} \approx 3.621320344$$

$$x=\frac{3-3\sqrt{2}}{2}\approx -0.621320344$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Me tāpiri te x^{2} ki ngā taha e rua.

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

Tangohia te 2x mai i ngā taha e rua.

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

Tangohia te 3 mai i ngā taha e rua.

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

Tangohia te 3 i te \frac{3}{4}, ka -\frac{9}{4}.

$$-x-\frac{9}{4}+x^{2}-2x=0$$

Pahekotia te -x me -2x, ka -3x.

$$-3x-\frac{9}{4}+x^{2}=0$$

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.

$$x^{2}-3x-\frac{9}{4}=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -3 mō b, me -\frac{9}{4} mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Pūrua -3.

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

Whakareatia -4 ki te -\frac{9}{4}.

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

Tāpiri 9 ki te 9.

$$x=\frac{-\left(-3\right)±\sqrt{18}}{2}$$

Tuhia te pūtakerua o te 18.

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

Ko te tauaro o -3 ko 3.

$$x=\frac{3±3\sqrt{2}}{2}$$

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

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

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

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

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Me tāpiri te x^{2} ki ngā taha e rua.

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

Tangohia te 2x mai i ngā taha e rua.

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

Tangohia te \frac{3}{4} mai i ngā taha e rua.

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

Tangohia te \frac{3}{4} i te 3, ka \frac{9}{4}.

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

Pahekotia te -x me -2x, ka -3x.

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

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.

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

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.

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

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

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

Tāpiri \frac{9}{4} ki te \frac{9}{4} 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.

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

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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{9}{2}$$

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

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

Whakarūnātia.

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

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

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