Whakaoti i te -(x+1)(x+4)-x+3*x=8

$$- ( x + 1 ) ( x + 4 ) - x + 3 \cdot x = 8$$

Whakaoti mō x (complex solution)

$$x=\frac{-\sqrt{39}i-3}{2}\approx -1.5-3.122498999i$$

$$x=\frac{-3+\sqrt{39}i}{2}\approx -1.5+3.122498999i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Hei kimi i te tauaro o x+1, kimihia te tauaro o ia taurangi.

$$\left(-x-1\right)\left(x+4\right)-x+3x=8$$

Me hoatu te āhuatanga tohatoha mā te whakarea ia tau o -x-1 ki ia tau o x+4.

$$-x^{2}-4x-x-4-x+3x=8$$

Pahekotia te -4x me -x, ka -5x.

$$-x^{2}-5x-4-x+3x=8$$

Pahekotia te -5x me -x, ka -6x.

$$-x^{2}-6x-4+3x=8$$

Pahekotia te -6x me 3x, ka -3x.

$$-x^{2}-3x-4=8$$

Tangohia te 8 mai i ngā taha e rua.

$$-x^{2}-3x-4-8=0$$

Tangohia te 8 i te -4, ka -12.

$$-x^{2}-3x-12=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -1 mō a, -3 mō b, me -12 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-1\right)\left(-12\right)}}{2\left(-1\right)}$$

Pūrua -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-1\right)\left(-12\right)}}{2\left(-1\right)}$$

Whakareatia -4 ki te -1.

$$x=\frac{-\left(-3\right)±\sqrt{9+4\left(-12\right)}}{2\left(-1\right)}$$

Whakareatia 4 ki te -12.

$$x=\frac{-\left(-3\right)±\sqrt{9-48}}{2\left(-1\right)}$$

Tāpiri 9 ki te -48.

$$x=\frac{-\left(-3\right)±\sqrt{-39}}{2\left(-1\right)}$$

Tuhia te pūtakerua o te -39.

$$x=\frac{-\left(-3\right)±\sqrt{39}i}{2\left(-1\right)}$$

Ko te tauaro o -3 ko 3.

$$x=\frac{3±\sqrt{39}i}{2\left(-1\right)}$$

Whakareatia 2 ki te -1.

$$x=\frac{3±\sqrt{39}i}{-2}$$

Nā, me whakaoti te whārite x=\frac{3±\sqrt{39}i}{-2} ina he tāpiri te ±. Tāpiri 3 ki te i\sqrt{39}.

$$x=\frac{3+\sqrt{39}i}{-2}$$

Whakawehe 3+i\sqrt{39} ki te -2.

$$x=\frac{-\sqrt{39}i-3}{2}$$

Nā, me whakaoti te whārite x=\frac{3±\sqrt{39}i}{-2} ina he tango te ±. Tango i\sqrt{39} mai i 3.

$$x=\frac{-\sqrt{39}i+3}{-2}$$

Whakawehe 3-i\sqrt{39} ki te -2.

$$x=\frac{-3+\sqrt{39}i}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{-\sqrt{39}i-3}{2} x=\frac{-3+\sqrt{39}i}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

Hei kimi i te tauaro o x+1, kimihia te tauaro o ia taurangi.

$$\left(-x-1\right)\left(x+4\right)-x+3x=8$$

Me hoatu te āhuatanga tohatoha mā te whakarea ia tau o -x-1 ki ia tau o x+4.

$$-x^{2}-4x-x-4-x+3x=8$$

Pahekotia te -4x me -x, ka -5x.

$$-x^{2}-5x-4-x+3x=8$$

Pahekotia te -5x me -x, ka -6x.

$$-x^{2}-6x-4+3x=8$$

Pahekotia te -6x me 3x, ka -3x.

$$-x^{2}-3x-4=8$$

Me tāpiri te 4 ki ngā taha e rua.

$$-x^{2}-3x=8+4$$

Tāpirihia te 8 ki te 4, ka 12.

$$-x^{2}-3x=12$$

Whakawehea ngā taha e rua ki te -1.

$$\frac{-x^{2}-3x}{-1}=\frac{12}{-1}$$

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

$$x^{2}+\left(-\frac{3}{-1}\right)x=\frac{12}{-1}$$

Whakawehe -3 ki te -1.

$$x^{2}+3x=\frac{12}{-1}$$

Whakawehe 12 ki te -1.

$$x^{2}+3x=-12$$

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.

$$x^{2}+3x+\left(\frac{3}{2}\right)^{2}=-12+\left(\frac{3}{2}\right)^{2}$$

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

$$x^{2}+3x+\frac{9}{4}=-12+\frac{9}{4}$$

Tāpiri -12 ki te \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=-\frac{39}{4}$$

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}.

$$\left(x+\frac{3}{2}\right)^{2}=-\frac{39}{4}$$

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

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{-\frac{39}{4}}$$

Whakarūnātia.

$$x+\frac{3}{2}=\frac{\sqrt{39}i}{2} x+\frac{3}{2}=-\frac{\sqrt{39}i}{2}$$

Me tango \frac{3}{2} mai i ngā taha e rua o te whārite.

$$x=\frac{-3+\sqrt{39}i}{2} x=\frac{-\sqrt{39}i-3}{2}$$