Whakaoti i te 3x+3=x^2+4x+5

$$3x+3= { x }^{ 2 } +4x+5$$

Whakaoti mō x (complex solution)

$$x=\frac{-\sqrt{7}i-1}{2}\approx -0.5-1.322875656i$$

$$x=\frac{-1+\sqrt{7}i}{2}\approx -0.5+1.322875656i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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

$$3x+3-x^{2}=4x+5$$

Tangohia te 4x mai i ngā taha e rua.

$$3x+3-x^{2}-4x=5$$

Pahekotia te 3x me -4x, ka -x.

$$-x+3-x^{2}=5$$

Tangohia te 5 mai i ngā taha e rua.

$$-x+3-x^{2}-5=0$$

Tangohia te 5 i te 3, ka -2.

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

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.

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

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

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

Whakareatia -4 ki te -1.

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

Whakareatia 4 ki te -2.

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

Tāpiri 1 ki te -8.

$$x=\frac{-\left(-1\right)±\sqrt{-7}}{2\left(-1\right)}$$

Tuhia te pūtakerua o te -7.

$$x=\frac{-\left(-1\right)±\sqrt{7}i}{2\left(-1\right)}$$

Ko te tauaro o -1 ko 1.

$$x=\frac{1±\sqrt{7}i}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

$$x=\frac{1±\sqrt{7}i}{-2}$$

Nā, me whakaoti te whārite x=\frac{1±\sqrt{7}i}{-2} ina he tāpiri te ±. Tāpiri 1 ki te i\sqrt{7}.

$$x=\frac{1+\sqrt{7}i}{-2}$$

Whakawehe 1+i\sqrt{7} ki te -2.

$$x=\frac{-\sqrt{7}i-1}{2}$$

Nā, me whakaoti te whārite x=\frac{1±\sqrt{7}i}{-2} ina he tango te ±. Tango i\sqrt{7} mai i 1.

$$x=\frac{-\sqrt{7}i+1}{-2}$$

Whakawehe 1-i\sqrt{7} ki te -2.

$$x=\frac{-1+\sqrt{7}i}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{-\sqrt{7}i-1}{2} x=\frac{-1+\sqrt{7}i}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

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

$$3x+3-x^{2}=4x+5$$

Tangohia te 4x mai i ngā taha e rua.

$$3x+3-x^{2}-4x=5$$

Pahekotia te 3x me -4x, ka -x.

$$-x+3-x^{2}=5$$

Tangohia te 3 mai i ngā taha e rua.

$$-x-x^{2}=5-3$$

Tangohia te 3 i te 5, ka 2.

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

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.

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

Whakawehea ngā taha e rua ki te -1.

$$\frac{-x^{2}-x}{-1}=\frac{2}{-1}$$

Mā te whakawehe ki te -1 ka wetekia te whakareanga ki te -1.

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

Whakawehe -1 ki te -1.

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

Whakawehe 2 ki te -1.

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

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

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

Pūruatia \frac{1}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+x+\frac{1}{4}=-2+\frac{1}{4}$$

Tāpiri -2 ki te \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=-\frac{7}{4}$$

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

$$\left(x+\frac{1}{2}\right)^{2}=-\frac{7}{4}$$

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

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{-\frac{7}{4}}$$

Whakarūnātia.

$$x+\frac{1}{2}=\frac{\sqrt{7}i}{2} x+\frac{1}{2}=-\frac{\sqrt{7}i}{2}$$

Me tango \frac{1}{2} mai i ngā taha e rua o te whārite.

$$x=\frac{-1+\sqrt{7}i}{2} x=\frac{-\sqrt{7}i-1}{2}$$