Whakaoti i te 5x^2-20x+20=20/9

$$5 { x }^{ 2 } -20x+20= \frac{ 20 }{ 9 }$$

Whakaoti mō x

$$x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

$$x = \frac{8}{3} = 2\frac{2}{3} \approx 2.666666667$$

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.

$$5x^{2}-20x+20=\frac{20}{9}$$

Me tango \frac{20}{9} mai i ngā taha e rua o te whārite.

$$5x^{2}-20x+20-\frac{20}{9}=\frac{20}{9}-\frac{20}{9}$$

Mā te tango i te \frac{20}{9} i a ia ake anō ka toe ko te 0.

$$5x^{2}-20x+20-\frac{20}{9}=0$$

Tango \frac{20}{9} mai i 20.

$$5x^{2}-20x+\frac{160}{9}=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 5 mō a, -20 mō b, me \frac{160}{9} mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 5\times \frac{160}{9}}}{2\times 5}$$

Pūrua -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 5\times \frac{160}{9}}}{2\times 5}$$

Whakareatia -4 ki te 5.

$$x=\frac{-\left(-20\right)±\sqrt{400-20\times \frac{160}{9}}}{2\times 5}$$

Whakareatia -20 ki te \frac{160}{9}.

$$x=\frac{-\left(-20\right)±\sqrt{400-\frac{3200}{9}}}{2\times 5}$$

Tāpiri 400 ki te -\frac{3200}{9}.

$$x=\frac{-\left(-20\right)±\sqrt{\frac{400}{9}}}{2\times 5}$$

Tuhia te pūtakerua o te \frac{400}{9}.

$$x=\frac{-\left(-20\right)±\frac{20}{3}}{2\times 5}$$

Ko te tauaro o -20 ko 20.

$$x=\frac{20±\frac{20}{3}}{2\times 5}$$

Whakareatia 2 ki te 5.

$$x=\frac{20±\frac{20}{3}}{10}$$

Nā, me whakaoti te whārite x=\frac{20±\frac{20}{3}}{10} ina he tāpiri te ±. Tāpiri 20 ki te \frac{20}{3}.

$$x=\frac{\frac{80}{3}}{10}$$

Whakawehe \frac{80}{3} ki te 10.

$$x=\frac{8}{3}$$

Nā, me whakaoti te whārite x=\frac{20±\frac{20}{3}}{10} ina he tango te ±. Tango \frac{20}{3} mai i 20.

$$x=\frac{\frac{40}{3}}{10}$$

Whakawehe \frac{40}{3} ki te 10.

$$x=\frac{4}{3}$$

Kua oti te whārite te whakatau.

$$x=\frac{8}{3} x=\frac{4}{3}$$

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.

$$5x^{2}-20x+20=\frac{20}{9}$$

Me tango 20 mai i ngā taha e rua o te whārite.

$$5x^{2}-20x+20-20=\frac{20}{9}-20$$

Mā te tango i te 20 i a ia ake anō ka toe ko te 0.

$$5x^{2}-20x=\frac{20}{9}-20$$

Tango 20 mai i \frac{20}{9}.

$$5x^{2}-20x=-\frac{160}{9}$$

Whakawehea ngā taha e rua ki te 5.

$$\frac{5x^{2}-20x}{5}=-\frac{\frac{160}{9}}{5}$$

Mā te whakawehe ki te 5 ka wetekia te whakareanga ki te 5.

$$x^{2}+\left(-\frac{20}{5}\right)x=-\frac{\frac{160}{9}}{5}$$

Whakawehe -20 ki te 5.

$$x^{2}-4x=-\frac{\frac{160}{9}}{5}$$

Whakawehe -\frac{160}{9} ki te 5.

$$x^{2}-4x=-\frac{32}{9}$$

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.

$$x^{2}-4x+\left(-2\right)^{2}=-\frac{32}{9}+\left(-2\right)^{2}$$

Pūrua -2.

$$x^{2}-4x+4=-\frac{32}{9}+4$$

Tāpiri -\frac{32}{9} ki te 4.

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

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}.

$$\left(x-2\right)^{2}=\frac{4}{9}$$

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

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{\frac{4}{9}}$$

Whakarūnātia.

$$x-2=\frac{2}{3} x-2=-\frac{2}{3}$$

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

$$x=\frac{8}{3} x=\frac{4}{3}$$