Whakaoti i te -2x^2-5x+5=0

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

Whakaoti mō x

$$x=\frac{\sqrt{65}-5}{4}\approx 0.765564437$$

$$x=\frac{-\sqrt{65}-5}{4}\approx -3.265564437$$

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.

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

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

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

Pūrua -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-2\right)\times 5}}{2\left(-2\right)}$$

Whakareatia -4 ki te -2.

$$x=\frac{-\left(-5\right)±\sqrt{25+8\times 5}}{2\left(-2\right)}$$

Whakareatia 8 ki te 5.

$$x=\frac{-\left(-5\right)±\sqrt{25+40}}{2\left(-2\right)}$$

Tāpiri 25 ki te 40.

$$x=\frac{-\left(-5\right)±\sqrt{65}}{2\left(-2\right)}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±\sqrt{65}}{2\left(-2\right)}$$

Whakareatia 2 ki te -2.

$$x=\frac{5±\sqrt{65}}{-4}$$

Nā, me whakaoti te whārite x=\frac{5±\sqrt{65}}{-4} ina he tāpiri te ±. Tāpiri 5 ki te \sqrt{65}.

$$x=\frac{\sqrt{65}+5}{-4}$$

Whakawehe 5+\sqrt{65} ki te -4.

$$x=\frac{-\sqrt{65}-5}{4}$$

Nā, me whakaoti te whārite x=\frac{5±\sqrt{65}}{-4} ina he tango te ±. Tango \sqrt{65} mai i 5.

$$x=\frac{5-\sqrt{65}}{-4}$$

Whakawehe 5-\sqrt{65} ki te -4.

$$x=\frac{\sqrt{65}-5}{4}$$

Kua oti te whārite te whakatau.

$$x=\frac{-\sqrt{65}-5}{4} x=\frac{\sqrt{65}-5}{4}$$

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.

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

Me tango 5 mai i ngā taha e rua o te whārite.

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

Mā te tango i te 5 i a ia ake anō ka toe ko te 0.

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

Whakawehea ngā taha e rua ki te -2.

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

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

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

Whakawehe -5 ki te -2.

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

Whakawehe -5 ki te -2.

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

Whakawehea te \frac{5}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{5}{4}. Nā, tāpiria te pūrua o te \frac{5}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+\frac{5}{2}x+\left(\frac{5}{4}\right)^{2}=\frac{5}{2}+\left(\frac{5}{4}\right)^{2}$$

Pūruatia \frac{5}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=\frac{5}{2}+\frac{25}{16}$$

Tāpiri \frac{5}{2} ki te \frac{25}{16} 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.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=\frac{65}{16}$$

Tauwehea x^{2}+\frac{5}{2}x+\frac{25}{16}. 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}.

$$\left(x+\frac{5}{4}\right)^{2}=\frac{65}{16}$$

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

$$\sqrt{\left(x+\frac{5}{4}\right)^{2}}=\sqrt{\frac{65}{16}}$$

Whakarūnātia.

$$x+\frac{5}{4}=\frac{\sqrt{65}}{4} x+\frac{5}{4}=-\frac{\sqrt{65}}{4}$$

Me tango \frac{5}{4} mai i ngā taha e rua o te whārite.

$$x=\frac{\sqrt{65}-5}{4} x=\frac{-\sqrt{65}-5}{4}$$