Whakaoti i te 20x=8*x^2/2

$$20x= \frac{ 8 \times { x }^{ 2 } }{ 2 }$$

Whakaoti mō x

$$x=5$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Whakareatia ngā taha e rua o te whārite ki te 2.

$$40x=8x^{2}$$

Tangohia te 8x^{2} mai i ngā taha e rua.

$$40x-8x^{2}=0$$

Tauwehea te x.

$$x\left(40-8x\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te 40-8x=0.

$$x=0 x=5$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakareatia ngā taha e rua o te whārite ki te 2.

$$40x=8x^{2}$$

Tangohia te 8x^{2} mai i ngā taha e rua.

$$40x-8x^{2}=0$$

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.

$$-8x^{2}+40x=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -8 mō a, 40 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-40±\sqrt{40^{2}}}{2\left(-8\right)}$$

Tuhia te pūtakerua o te 40^{2}.

$$x=\frac{-40±40}{2\left(-8\right)}$$

Whakareatia 2 ki te -8.

$$x=\frac{-40±40}{-16}$$

Nā, me whakaoti te whārite x=\frac{-40±40}{-16} ina he tāpiri te ±. Tāpiri -40 ki te 40.

$$x=\frac{0}{-16}$$

Whakawehe 0 ki te -16.

$$x=0$$

Nā, me whakaoti te whārite x=\frac{-40±40}{-16} ina he tango te ±. Tango 40 mai i -40.

$$x=-\frac{80}{-16}$$

Whakawehe -80 ki te -16.

$$x=5$$

Kua oti te whārite te whakatau.

$$x=0 x=5$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakareatia ngā taha e rua o te whārite ki te 2.

$$40x=8x^{2}$$

Tangohia te 8x^{2} mai i ngā taha e rua.

$$40x-8x^{2}=0$$

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.

$$-8x^{2}+40x=0$$

Whakawehea ngā taha e rua ki te -8.

$$\frac{-8x^{2}+40x}{-8}=\frac{0}{-8}$$

Mā te whakawehe ki te -8 ka wetekia te whakareanga ki te -8.

$$x^{2}+\frac{40}{-8}x=\frac{0}{-8}$$

Whakawehe 40 ki te -8.

$$x^{2}-5x=\frac{0}{-8}$$

Whakawehe 0 ki te -8.

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

Whakawehea te -5, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{5}{2}. Nā, tāpiria te pūrua o te -\frac{5}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

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

Pūruatia -\frac{5}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-5x+\frac{25}{4}=\frac{25}{4}$$

Tauwehea x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{25}{4}$$

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

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Whakarūnātia.

$$x-\frac{5}{2}=\frac{5}{2} x-\frac{5}{2}=-\frac{5}{2}$$

Me tāpiri \frac{5}{2} ki ngā taha e rua o te whārite.

$$x=5 x=0$$