Whakaoti i te 5x^2-8x+3(x+2)x=0

$$5 x ^ { 2 } - 8 x + 3 ( x + 2 ) x = 0$$

Whakaoti mō x

$$x=\frac{1}{4}=0.25$$

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Whakamahia te āhuatanga tohatoha hei whakarea te 3 ki te x+2.

$$5x^{2}-8x+\left(3x+6\right)x=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te 3x+6 ki te x.

$$5x^{2}-8x+3x^{2}+6x=0$$

Pahekotia te 5x^{2} me 3x^{2}, ka 8x^{2}.

$$8x^{2}-8x+6x=0$$

Pahekotia te -8x me 6x, ka -2x.

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

Tauwehea te x.

$$x\left(8x-2\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te 8x-2=0.

$$x=0 x=\frac{1}{4}$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 3 ki te x+2.

$$5x^{2}-8x+\left(3x+6\right)x=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te 3x+6 ki te x.

$$5x^{2}-8x+3x^{2}+6x=0$$

Pahekotia te 5x^{2} me 3x^{2}, ka 8x^{2}.

$$8x^{2}-8x+6x=0$$

Pahekotia te -8x me 6x, ka -2x.

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

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

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

Tuhia te pūtakerua o te \left(-2\right)^{2}.

$$x=\frac{-\left(-2\right)±2}{2\times 8}$$

Ko te tauaro o -2 ko 2.

$$x=\frac{2±2}{2\times 8}$$

Whakareatia 2 ki te 8.

$$x=\frac{2±2}{16}$$

Nā, me whakaoti te whārite x=\frac{2±2}{16} ina he tāpiri te ±. Tāpiri 2 ki te 2.

$$x=\frac{4}{16}$$

Whakahekea te hautanga \frac{4}{16} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 4.

$$x=\frac{1}{4}$$

Nā, me whakaoti te whārite x=\frac{2±2}{16} ina he tango te ±. Tango 2 mai i 2.

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

Whakawehe 0 ki te 16.

$$x=0$$

Kua oti te whārite te whakatau.

$$x=\frac{1}{4} x=0$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tohatoha hei whakarea te 3 ki te x+2.

$$5x^{2}-8x+\left(3x+6\right)x=0$$

Whakamahia te āhuatanga tohatoha hei whakarea te 3x+6 ki te x.

$$5x^{2}-8x+3x^{2}+6x=0$$

Pahekotia te 5x^{2} me 3x^{2}, ka 8x^{2}.

$$8x^{2}-8x+6x=0$$

Pahekotia te -8x me 6x, ka -2x.

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

Whakawehea ngā taha e rua ki te 8.

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

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

$$x^{2}+\left(-\frac{2}{8}\right)x=\frac{0}{8}$$

Whakahekea te hautanga \frac{-2}{8} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$x^{2}-\frac{1}{4}x=\frac{0}{8}$$

Whakawehe 0 ki te 8.

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

Whakawehea te -\frac{1}{4}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{8}. Nā, tāpiria te pūrua o te -\frac{1}{8} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{1}{4}x+\left(-\frac{1}{8}\right)^{2}=\left(-\frac{1}{8}\right)^{2}$$

Pūruatia -\frac{1}{8} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{1}{4}x+\frac{1}{64}=\frac{1}{64}$$

Tauwehea x^{2}-\frac{1}{4}x+\frac{1}{64}. 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}.

$$\left(x-\frac{1}{8}\right)^{2}=\frac{1}{64}$$

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

$$\sqrt{\left(x-\frac{1}{8}\right)^{2}}=\sqrt{\frac{1}{64}}$$

Whakarūnātia.

$$x-\frac{1}{8}=\frac{1}{8} x-\frac{1}{8}=-\frac{1}{8}$$

Me tāpiri \frac{1}{8} ki ngā taha e rua o te whārite.

$$x=\frac{1}{4} x=0$$