Whakaoti i te 5x(3-4x)=15x-4x

$$5 x ( 3 - 4 x ) = 15 x - 4 x$$

Whakaoti mō x

$$x=\frac{1}{5}=0.2$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Whakamahia te āhuatanga tohatoha hei whakarea te 5x ki te 3-4x.

$$15x-20x^{2}=15x-4x$$

Pahekotia te 15x me -4x, ka 11x.

$$15x-20x^{2}=11x$$

Tangohia te 11x mai i ngā taha e rua.

$$15x-20x^{2}-11x=0$$

Pahekotia te 15x me -11x, ka 4x.

$$4x-20x^{2}=0$$

Tauwehea te x.

$$x\left(4-20x\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te 4-20x=0.

$$x=0 x=\frac{1}{5}$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 5x ki te 3-4x.

$$15x-20x^{2}=15x-4x$$

Pahekotia te 15x me -4x, ka 11x.

$$15x-20x^{2}=11x$$

Tangohia te 11x mai i ngā taha e rua.

$$15x-20x^{2}-11x=0$$

Pahekotia te 15x me -11x, ka 4x.

$$4x-20x^{2}=0$$

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.

$$-20x^{2}+4x=0$$

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

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

Tuhia te pūtakerua o te 4^{2}.

$$x=\frac{-4±4}{2\left(-20\right)}$$

Whakareatia 2 ki te -20.

$$x=\frac{-4±4}{-40}$$

Nā, me whakaoti te whārite x=\frac{-4±4}{-40} ina he tāpiri te ±. Tāpiri -4 ki te 4.

$$x=\frac{0}{-40}$$

Whakawehe 0 ki te -40.

$$x=0$$

Nā, me whakaoti te whārite x=\frac{-4±4}{-40} ina he tango te ±. Tango 4 mai i -4.

$$x=-\frac{8}{-40}$$

Whakahekea te hautanga \frac{-8}{-40} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 8.

$$x=\frac{1}{5}$$

Kua oti te whārite te whakatau.

$$x=0 x=\frac{1}{5}$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 5x ki te 3-4x.

$$15x-20x^{2}=15x-4x$$

Pahekotia te 15x me -4x, ka 11x.

$$15x-20x^{2}=11x$$

Tangohia te 11x mai i ngā taha e rua.

$$15x-20x^{2}-11x=0$$

Pahekotia te 15x me -11x, ka 4x.

$$4x-20x^{2}=0$$

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.

$$-20x^{2}+4x=0$$

Whakawehea ngā taha e rua ki te -20.

$$\frac{-20x^{2}+4x}{-20}=\frac{0}{-20}$$

Mā te whakawehe ki te -20 ka wetekia te whakareanga ki te -20.

$$x^{2}+\frac{4}{-20}x=\frac{0}{-20}$$

Whakahekea te hautanga \frac{4}{-20} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$x^{2}-\frac{1}{5}x=\frac{0}{-20}$$

Whakawehe 0 ki te -20.

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

Whakawehea te -\frac{1}{5}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{10}. Nā, tāpiria te pūrua o te -\frac{1}{10} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{1}{5}x+\left(-\frac{1}{10}\right)^{2}=\left(-\frac{1}{10}\right)^{2}$$

Pūruatia -\frac{1}{10} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{1}{100}$$

Tauwehea x^{2}-\frac{1}{5}x+\frac{1}{100}. 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}.

$$\left(x-\frac{1}{10}\right)^{2}=\frac{1}{100}$$

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

$$\sqrt{\left(x-\frac{1}{10}\right)^{2}}=\sqrt{\frac{1}{100}}$$

Whakarūnātia.

$$x-\frac{1}{10}=\frac{1}{10} x-\frac{1}{10}=-\frac{1}{10}$$

Me tāpiri \frac{1}{10} ki ngā taha e rua o te whārite.

$$x=\frac{1}{5} x=0$$