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

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

Whakaoti mō x

$$x = \frac{\sqrt{11} - 1}{2} \approx 1.158312395$$

$$x=\frac{-\sqrt{11}-1}{2}\approx -2.158312395$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 3x^{2} mai i ngā taha e rua.

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

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

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

Tangohia te 5 mai i ngā taha e rua.

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

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

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

Pūrua 2.

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

Whakareatia -4 ki te 2.

$$x=\frac{-2±\sqrt{4-8\left(-5\right)}}{2\times 2}$$

Whakareatia -8 ki te -5.

$$x=\frac{-2±\sqrt{4+40}}{2\times 2}$$

Tāpiri 4 ki te 40.

$$x=\frac{-2±\sqrt{44}}{2\times 2}$$

Tuhia te pūtakerua o te 44.

$$x=\frac{-2±2\sqrt{11}}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{-2±2\sqrt{11}}{4}$$

Nā, me whakaoti te whārite x=\frac{-2±2\sqrt{11}}{4} ina he tāpiri te ±. Tāpiri -2 ki te 2\sqrt{11}.

$$x=\frac{2\sqrt{11}-2}{4}$$

Whakawehe -2+2\sqrt{11} ki te 4.

$$x=\frac{\sqrt{11}-1}{2}$$

Nā, me whakaoti te whārite x=\frac{-2±2\sqrt{11}}{4} ina he tango te ±. Tango 2\sqrt{11} mai i -2.

$$x=\frac{-2\sqrt{11}-2}{4}$$

Whakawehe -2-2\sqrt{11} ki te 4.

$$x=\frac{-\sqrt{11}-1}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{11}-1}{2} x=\frac{-\sqrt{11}-1}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 3x^{2} mai i ngā taha e rua.

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

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

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

Whakawehea ngā taha e rua ki te 2.

$$\frac{2x^{2}+2x}{2}=\frac{5}{2}$$

Mā te whakawehe ki te 2 ka wetekia te whakareanga ki te 2.

$$x^{2}+\frac{2}{2}x=\frac{5}{2}$$

Whakawehe 2 ki te 2.

$$x^{2}+x=\frac{5}{2}$$

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

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

Pūruatia \frac{1}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+x+\frac{1}{4}=\frac{5}{2}+\frac{1}{4}$$

Tāpiri \frac{5}{2} ki te \frac{1}{4} mā te kimi i te tauraro pātahi me te tāpiri i ngā taurunga. Ka whakaiti i te hautanga ki ngā kīanga tau iti rawa e taea ana.

$$x^{2}+x+\frac{1}{4}=\frac{11}{4}$$

Tauwehea x^{2}+x+\frac{1}{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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{11}{4}$$

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

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{11}{4}}$$

Whakarūnātia.

$$x+\frac{1}{2}=\frac{\sqrt{11}}{2} x+\frac{1}{2}=-\frac{\sqrt{11}}{2}$$

Me tango \frac{1}{2} mai i ngā taha e rua o te whārite.

$$x=\frac{\sqrt{11}-1}{2} x=\frac{-\sqrt{11}-1}{2}$$