Whakaoti i te x^2/4-x+5=0

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

Whakaoti mō x (complex solution)

$$x=2+4i$$

$$x=2-4i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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.

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

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi \frac{1}{4} mō a, -1 mō b, me 5 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times \frac{1}{4}\times 5}}{2\times \frac{1}{4}}$$

Whakareatia -4 ki te \frac{1}{4}.

$$x=\frac{-\left(-1\right)±\sqrt{1-5}}{2\times \frac{1}{4}}$$

Tāpiri 1 ki te -5.

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

Tuhia te pūtakerua o te -4.

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

Ko te tauaro o -1 ko 1.

$$x=\frac{1±2i}{2\times \frac{1}{4}}$$

Whakareatia 2 ki te \frac{1}{4}.

$$x=\frac{1±2i}{\frac{1}{2}}$$

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

$$x=\frac{1+2i}{\frac{1}{2}}$$

Whakawehe 1+2i ki te \frac{1}{2} mā te whakarea 1+2i ki te tau huripoki o \frac{1}{2}.

$$x=2+4i$$

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

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

Whakawehe 1-2i ki te \frac{1}{2} mā te whakarea 1-2i ki te tau huripoki o \frac{1}{2}.

$$x=2-4i$$

Kua oti te whārite te whakatau.

$$x=2+4i x=2-4i$$

Ngā Upane mō te Whakaoti i te Pūrua

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.

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

Me tango 5 mai i ngā taha e rua o te whārite.

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

Mā te tango i te 5 i a ia ake anō ka toe ko te 0.

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

Me whakarea ngā taha e rua ki te 4.

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

Mā te whakawehe ki te \frac{1}{4} ka wetekia te whakareanga ki te \frac{1}{4}.

$$x^{2}+\left(-\frac{1}{\frac{1}{4}}\right)x=-\frac{5}{\frac{1}{4}}$$

Whakawehe -1 ki te \frac{1}{4} mā te whakarea -1 ki te tau huripoki o \frac{1}{4}.

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

Whakawehe -5 ki te \frac{1}{4} mā te whakarea -5 ki te tau huripoki o \frac{1}{4}.

$$x^{2}-4x=-20$$

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

$$x^{2}-4x+\left(-2\right)^{2}=-20+\left(-2\right)^{2}$$

Pūrua -2.

$$x^{2}-4x+4=-20+4$$

Tāpiri -20 ki te 4.

$$x^{2}-4x+4=-16$$

Tauwehea x^{2}-4x+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}.

$$\left(x-2\right)^{2}=-16$$

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

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{-16}$$

Whakarūnātia.

$$x-2=4i x-2=-4i$$

Me tāpiri 2 ki ngā taha e rua o te whārite.

$$x=2+4i x=2-4i$$