Whakaoti i te -2x^2+6x+1=0

$$-2 { x }^{ 2 } +6x+1 = 0$$

Whakaoti mō x

$$x = \frac{\sqrt{11} + 3}{2} \approx 3.158312395$$

$$x=\frac{3-\sqrt{11}}{2}\approx -0.158312395$$

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.

$$-2x^{2}+6x+1=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi -2 mō a, 6 mō b, me 1 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-2\right)}}{2\left(-2\right)}$$

Pūrua 6.

$$x=\frac{-6±\sqrt{36-4\left(-2\right)}}{2\left(-2\right)}$$

Whakareatia -4 ki te -2.

$$x=\frac{-6±\sqrt{36+8}}{2\left(-2\right)}$$

Tāpiri 36 ki te 8.

$$x=\frac{-6±\sqrt{44}}{2\left(-2\right)}$$

Tuhia te pūtakerua o te 44.

$$x=\frac{-6±2\sqrt{11}}{2\left(-2\right)}$$

Whakareatia 2 ki te -2.

$$x=\frac{-6±2\sqrt{11}}{-4}$$

Nā, me whakaoti te whārite x=\frac{-6±2\sqrt{11}}{-4} ina he tāpiri te ±. Tāpiri -6 ki te 2\sqrt{11}.

$$x=\frac{2\sqrt{11}-6}{-4}$$

Whakawehe -6+2\sqrt{11} ki te -4.

$$x=\frac{3-\sqrt{11}}{2}$$

Nā, me whakaoti te whārite x=\frac{-6±2\sqrt{11}}{-4} ina he tango te ±. Tango 2\sqrt{11} mai i -6.

$$x=\frac{-2\sqrt{11}-6}{-4}$$

Whakawehe -6-2\sqrt{11} ki te -4.

$$x=\frac{\sqrt{11}+3}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{3-\sqrt{11}}{2} x=\frac{\sqrt{11}+3}{2}$$

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.

$$-2x^{2}+6x+1=0$$

Me tango 1 mai i ngā taha e rua o te whārite.

$$-2x^{2}+6x+1-1=-1$$

Mā te tango i te 1 i a ia ake anō ka toe ko te 0.

$$-2x^{2}+6x=-1$$

Whakawehea ngā taha e rua ki te -2.

$$\frac{-2x^{2}+6x}{-2}=-\frac{1}{-2}$$

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

$$x^{2}+\frac{6}{-2}x=-\frac{1}{-2}$$

Whakawehe 6 ki te -2.

$$x^{2}-3x=-\frac{1}{-2}$$

Whakawehe -1 ki te -2.

$$x^{2}-3x=\frac{1}{2}$$

Whakawehea te -3, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{3}{2}. Nā, tāpiria te pūrua o te -\frac{3}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=\frac{1}{2}+\left(-\frac{3}{2}\right)^{2}$$

Pūruatia -\frac{3}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-3x+\frac{9}{4}=\frac{1}{2}+\frac{9}{4}$$

Tāpiri \frac{1}{2} ki te \frac{9}{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.

$$x^{2}-3x+\frac{9}{4}=\frac{11}{4}$$

Tauwehea x^{2}-3x+\frac{9}{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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{11}{4}$$

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

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{11}{4}}$$

Whakarūnātia.

$$x-\frac{3}{2}=\frac{\sqrt{11}}{2} x-\frac{3}{2}=-\frac{\sqrt{11}}{2}$$

Me tāpiri \frac{3}{2} ki ngā taha e rua o te whārite.

$$x=\frac{\sqrt{11}+3}{2} x=\frac{3-\sqrt{11}}{2}$$