Whakaoti i te 2x^2-15x-1=0

$$2 x ^ { 2 } - 15 x - 1 = 0$$

Whakaoti mō x

$$x = \frac{\sqrt{233} + 15}{4} \approx 7.566084381$$

$$x=\frac{15-\sqrt{233}}{4}\approx -0.066084381$$

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.

$$2x^{2}-15x-1=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 2 mō a, -15 mō b, me -1 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\times 2\left(-1\right)}}{2\times 2}$$

Pūrua -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\times 2\left(-1\right)}}{2\times 2}$$

Whakareatia -4 ki te 2.

$$x=\frac{-\left(-15\right)±\sqrt{225-8\left(-1\right)}}{2\times 2}$$

Whakareatia -8 ki te -1.

$$x=\frac{-\left(-15\right)±\sqrt{225+8}}{2\times 2}$$

Tāpiri 225 ki te 8.

$$x=\frac{-\left(-15\right)±\sqrt{233}}{2\times 2}$$

Ko te tauaro o -15 ko 15.

$$x=\frac{15±\sqrt{233}}{2\times 2}$$

Whakareatia 2 ki te 2.

$$x=\frac{15±\sqrt{233}}{4}$$

Nā, me whakaoti te whārite x=\frac{15±\sqrt{233}}{4} ina he tāpiri te ±. Tāpiri 15 ki te \sqrt{233}.

$$x=\frac{\sqrt{233}+15}{4}$$

Nā, me whakaoti te whārite x=\frac{15±\sqrt{233}}{4} ina he tango te ±. Tango \sqrt{233} mai i 15.

$$x=\frac{15-\sqrt{233}}{4}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{233}+15}{4} x=\frac{15-\sqrt{233}}{4}$$

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.

$$2x^{2}-15x-1=0$$

Me tāpiri 1 ki ngā taha e rua o te whārite.

$$2x^{2}-15x-1-\left(-1\right)=-\left(-1\right)$$

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

$$2x^{2}-15x=-\left(-1\right)$$

Tango -1 mai i 0.

$$2x^{2}-15x=1$$

Whakawehea ngā taha e rua ki te 2.

$$\frac{2x^{2}-15x}{2}=\frac{1}{2}$$

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

$$x^{2}-\frac{15}{2}x=\frac{1}{2}$$

Whakawehea te -\frac{15}{2}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{15}{4}. Nā, tāpiria te pūrua o te -\frac{15}{4} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{15}{2}x+\left(-\frac{15}{4}\right)^{2}=\frac{1}{2}+\left(-\frac{15}{4}\right)^{2}$$

Pūruatia -\frac{15}{4} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{15}{2}x+\frac{225}{16}=\frac{1}{2}+\frac{225}{16}$$

Tāpiri \frac{1}{2} ki te \frac{225}{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.

$$x^{2}-\frac{15}{2}x+\frac{225}{16}=\frac{233}{16}$$

Tauwehea x^{2}-\frac{15}{2}x+\frac{225}{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}.

$$\left(x-\frac{15}{4}\right)^{2}=\frac{233}{16}$$

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

$$\sqrt{\left(x-\frac{15}{4}\right)^{2}}=\sqrt{\frac{233}{16}}$$

Whakarūnātia.

$$x-\frac{15}{4}=\frac{\sqrt{233}}{4} x-\frac{15}{4}=-\frac{\sqrt{233}}{4}$$

Me tāpiri \frac{15}{4} ki ngā taha e rua o te whārite.

$$x=\frac{\sqrt{233}+15}{4} x=\frac{15-\sqrt{233}}{4}$$