Whakaoti i te 2x+1=4x^2+4x+5

$$2x+1=4 { x }^{ 2 } +4x+5$$

Whakaoti mō x (complex solution)

$$x=\frac{-1+\sqrt{15}i}{4}\approx -0.25+0.968245837i$$

$$x=\frac{-\sqrt{15}i-1}{4}\approx -0.25-0.968245837i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 4x^{2} mai i ngā taha e rua.

$$2x+1-4x^{2}=4x+5$$

Tangohia te 4x mai i ngā taha e rua.

$$2x+1-4x^{2}-4x=5$$

Pahekotia te 2x me -4x, ka -2x.

$$-2x+1-4x^{2}=5$$

Tangohia te 5 mai i ngā taha e rua.

$$-2x+1-4x^{2}-5=0$$

Tangohia te 5 i te 1, ka -4.

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

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.

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

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

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

Pūrua -2.

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

Whakareatia -4 ki te -4.

$$x=\frac{-\left(-2\right)±\sqrt{4+16\left(-4\right)}}{2\left(-4\right)}$$

Whakareatia 16 ki te -4.

$$x=\frac{-\left(-2\right)±\sqrt{4-64}}{2\left(-4\right)}$$

Tāpiri 4 ki te -64.

$$x=\frac{-\left(-2\right)±\sqrt{-60}}{2\left(-4\right)}$$

Tuhia te pūtakerua o te -60.

$$x=\frac{-\left(-2\right)±2\sqrt{15}i}{2\left(-4\right)}$$

Ko te tauaro o -2 ko 2.

$$x=\frac{2±2\sqrt{15}i}{2\left(-4\right)}$$

Whakareatia 2 ki te -4.

$$x=\frac{2±2\sqrt{15}i}{-8}$$

Nā, me whakaoti te whārite x=\frac{2±2\sqrt{15}i}{-8} ina he tāpiri te ±. Tāpiri 2 ki te 2i\sqrt{15}.

$$x=\frac{2+2\sqrt{15}i}{-8}$$

Whakawehe 2+2i\sqrt{15} ki te -8.

$$x=\frac{-\sqrt{15}i-1}{4}$$

Nā, me whakaoti te whārite x=\frac{2±2\sqrt{15}i}{-8} ina he tango te ±. Tango 2i\sqrt{15} mai i 2.

$$x=\frac{-2\sqrt{15}i+2}{-8}$$

Whakawehe 2-2i\sqrt{15} ki te -8.

$$x=\frac{-1+\sqrt{15}i}{4}$$

Kua oti te whārite te whakatau.

$$x=\frac{-\sqrt{15}i-1}{4} x=\frac{-1+\sqrt{15}i}{4}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 4x^{2} mai i ngā taha e rua.

$$2x+1-4x^{2}=4x+5$$

Tangohia te 4x mai i ngā taha e rua.

$$2x+1-4x^{2}-4x=5$$

Pahekotia te 2x me -4x, ka -2x.

$$-2x+1-4x^{2}=5$$

Tangohia te 1 mai i ngā taha e rua.

$$-2x-4x^{2}=5-1$$

Tangohia te 1 i te 5, ka 4.

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

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.

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

Whakawehea ngā taha e rua ki te -4.

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

Mā te whakawehe ki te -4 ka wetekia te whakareanga ki te -4.

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

Whakahekea te hautanga \frac{-2}{-4} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

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

Whakawehe 4 ki te -4.

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

Whakawehea te \frac{1}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{1}{4}. Nā, tāpiria te pūrua o te \frac{1}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia \frac{1}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

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

Tāpiri -1 ki te \frac{1}{16}.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=-\frac{15}{16}$$

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

$$\left(x+\frac{1}{4}\right)^{2}=-\frac{15}{16}$$

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

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{-\frac{15}{16}}$$

Whakarūnātia.

$$x+\frac{1}{4}=\frac{\sqrt{15}i}{4} x+\frac{1}{4}=-\frac{\sqrt{15}i}{4}$$

Me tango \frac{1}{4} mai i ngā taha e rua o te whārite.

$$x=\frac{-1+\sqrt{15}i}{4} x=\frac{-\sqrt{15}i-1}{4}$$