Whakaoti i te (18-3x)x=40

$$(18-3x)x=40$$

Whakaoti mō x (complex solution)

$$x=-\frac{\sqrt{39}i}{3}+3\approx 3-2.081665999i$$

$$x=\frac{\sqrt{39}i}{3}+3\approx 3+2.081665999i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 18-3x ki te x.

$$18x-3x^{2}=40$$

Tangohia te 40 mai i ngā taha e rua.

$$18x-3x^{2}-40=0$$

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.

$$-3x^{2}+18x-40=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -3 mō a, 18 mō b, me -40 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-18±\sqrt{18^{2}-4\left(-3\right)\left(-40\right)}}{2\left(-3\right)}$$

Pūrua 18.

$$x=\frac{-18±\sqrt{324-4\left(-3\right)\left(-40\right)}}{2\left(-3\right)}$$

Whakareatia -4 ki te -3.

$$x=\frac{-18±\sqrt{324+12\left(-40\right)}}{2\left(-3\right)}$$

Whakareatia 12 ki te -40.

$$x=\frac{-18±\sqrt{324-480}}{2\left(-3\right)}$$

Tāpiri 324 ki te -480.

$$x=\frac{-18±\sqrt{-156}}{2\left(-3\right)}$$

Tuhia te pūtakerua o te -156.

$$x=\frac{-18±2\sqrt{39}i}{2\left(-3\right)}$$

Whakareatia 2 ki te -3.

$$x=\frac{-18±2\sqrt{39}i}{-6}$$

Nā, me whakaoti te whārite x=\frac{-18±2\sqrt{39}i}{-6} ina he tāpiri te ±. Tāpiri -18 ki te 2i\sqrt{39}.

$$x=\frac{-18+2\sqrt{39}i}{-6}$$

Whakawehe -18+2i\sqrt{39} ki te -6.

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

Nā, me whakaoti te whārite x=\frac{-18±2\sqrt{39}i}{-6} ina he tango te ±. Tango 2i\sqrt{39} mai i -18.

$$x=\frac{-2\sqrt{39}i-18}{-6}$$

Whakawehe -18-2i\sqrt{39} ki te -6.

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

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 18-3x ki te x.

$$18x-3x^{2}=40$$

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.

$$-3x^{2}+18x=40$$

Whakawehea ngā taha e rua ki te -3.

$$\frac{-3x^{2}+18x}{-3}=\frac{40}{-3}$$

Mā te whakawehe ki te -3 ka wetekia te whakareanga ki te -3.

$$x^{2}+\frac{18}{-3}x=\frac{40}{-3}$$

Whakawehe 18 ki te -3.

$$x^{2}-6x=\frac{40}{-3}$$

Whakawehe 40 ki te -3.

$$x^{2}-6x=-\frac{40}{3}$$

Whakawehea te -6, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -3. Nā, tāpiria te pūrua o te -3 ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-6x+\left(-3\right)^{2}=-\frac{40}{3}+\left(-3\right)^{2}$$

Pūrua -3.

$$x^{2}-6x+9=-\frac{40}{3}+9$$

Tāpiri -\frac{40}{3} ki te 9.

$$x^{2}-6x+9=-\frac{13}{3}$$

Tauwehea x^{2}-6x+9. 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}.

$$\left(x-3\right)^{2}=-\frac{13}{3}$$

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

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{-\frac{13}{3}}$$

Whakarūnātia.

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

Me tāpiri 3 ki ngā taha e rua o te whārite.

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