Whakaoti i te 5x^2+2x=1x^2+3x

$$5 x ^ { 2 } + 2 x = 1 x ^ { 2 } + 3 x$$

Whakaoti mō x

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

$$x=0$$

Ngā Upane Whakamahi Tauwehe

Tangohia te 1x^{2} mai i ngā taha e rua.

$$5x^{2}+2x-x^{2}=3x$$

Pahekotia te 5x^{2} me -x^{2}, ka 4x^{2}.

$$4x^{2}+2x=3x$$

Tangohia te 3x mai i ngā taha e rua.

$$4x^{2}+2x-3x=0$$

Pahekotia te 2x me -3x, ka -x.

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

Tauwehea te x.

$$x\left(4x-1\right)=0$$

Hei kimi otinga whārite, me whakaoti te x=0 me te 4x-1=0.

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

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 1x^{2} mai i ngā taha e rua.

$$5x^{2}+2x-x^{2}=3x$$

Pahekotia te 5x^{2} me -x^{2}, ka 4x^{2}.

$$4x^{2}+2x=3x$$

Tangohia te 3x mai i ngā taha e rua.

$$4x^{2}+2x-3x=0$$

Pahekotia te 2x me -3x, ka -x.

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

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 4 mō a, -1 mō b, me 0 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1}}{2\times 4}$$

Tuhia te pūtakerua o te 1.

$$x=\frac{-\left(-1\right)±1}{2\times 4}$$

Ko te tauaro o -1 ko 1.

$$x=\frac{1±1}{2\times 4}$$

Whakareatia 2 ki te 4.

$$x=\frac{1±1}{8}$$

Nā, me whakaoti te whārite x=\frac{1±1}{8} ina he tāpiri te ±. Tāpiri 1 ki te 1.

$$x=\frac{2}{8}$$

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

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

Nā, me whakaoti te whārite x=\frac{1±1}{8} ina he tango te ±. Tango 1 mai i 1.

$$x=\frac{0}{8}$$

Whakawehe 0 ki te 8.

$$x=0$$

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 1x^{2} mai i ngā taha e rua.

$$5x^{2}+2x-x^{2}=3x$$

Pahekotia te 5x^{2} me -x^{2}, ka 4x^{2}.

$$4x^{2}+2x=3x$$

Tangohia te 3x mai i ngā taha e rua.

$$4x^{2}+2x-3x=0$$

Pahekotia te 2x me -3x, ka -x.

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

Whakawehea ngā taha e rua ki te 4.

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

Mā te whakawehe ki te 4 ka wetekia te whakareanga ki te 4.

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

Whakawehe 0 ki te 4.

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

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 te x^{2}-\frac{1}{4}x+\frac{1}{64}. Ko te tikanga, ina ko x^{2}+bx+c he pūrua tika, ka taea te tauwehe 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.

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