Whakaoti i te 7=15x^2-5x

$$7 = 15 x ^ { 2 } - 5 x$$

Whakaoti mō x

$$x=\frac{\sqrt{445}}{30}+\frac{1}{6}\approx 0.869834104$$

$$x=-\frac{\sqrt{445}}{30}+\frac{1}{6}\approx -0.53650077$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$15x^{2}-5x=7$$

Tangohia te 7 mai i ngā taha e rua.

$$15x^{2}-5x-7=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 15 mō a, -5 mō b, me -7 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 15\left(-7\right)}}{2\times 15}$$

Pūrua -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 15\left(-7\right)}}{2\times 15}$$

Whakareatia -4 ki te 15.

$$x=\frac{-\left(-5\right)±\sqrt{25-60\left(-7\right)}}{2\times 15}$$

Whakareatia -60 ki te -7.

$$x=\frac{-\left(-5\right)±\sqrt{25+420}}{2\times 15}$$

Tāpiri 25 ki te 420.

$$x=\frac{-\left(-5\right)±\sqrt{445}}{2\times 15}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±\sqrt{445}}{2\times 15}$$

Whakareatia 2 ki te 15.

$$x=\frac{5±\sqrt{445}}{30}$$

Nā, me whakaoti te whārite x=\frac{5±\sqrt{445}}{30} ina he tāpiri te ±. Tāpiri 5 ki te \sqrt{445}.

$$x=\frac{\sqrt{445}+5}{30}$$

Whakawehe 5+\sqrt{445} ki te 30.

$$x=\frac{\sqrt{445}}{30}+\frac{1}{6}$$

Nā, me whakaoti te whārite x=\frac{5±\sqrt{445}}{30} ina he tango te ±. Tango \sqrt{445} mai i 5.

$$x=\frac{5-\sqrt{445}}{30}$$

Whakawehe 5-\sqrt{445} ki te 30.

$$x=-\frac{\sqrt{445}}{30}+\frac{1}{6}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{445}}{30}+\frac{1}{6} x=-\frac{\sqrt{445}}{30}+\frac{1}{6}$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$15x^{2}-5x=7$$

Whakawehea ngā taha e rua ki te 15.

$$\frac{15x^{2}-5x}{15}=\frac{7}{15}$$

Mā te whakawehe ki te 15 ka wetekia te whakareanga ki te 15.

$$x^{2}+\left(-\frac{5}{15}\right)x=\frac{7}{15}$$

Whakahekea te hautanga \frac{-5}{15} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 5.

$$x^{2}-\frac{1}{3}x=\frac{7}{15}$$

Whakawehea te -\frac{1}{3}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{1}{6}. Nā, tāpiria te pūrua o te -\frac{1}{6} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=\frac{7}{15}+\left(-\frac{1}{6}\right)^{2}$$

Pūruatia -\frac{1}{6} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{7}{15}+\frac{1}{36}$$

Tāpiri \frac{7}{15} ki te \frac{1}{36} 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.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{89}{180}$$

Tauwehea x^{2}-\frac{1}{3}x+\frac{1}{36}. 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}.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{89}{180}$$

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

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{89}{180}}$$

Whakarūnātia.

$$x-\frac{1}{6}=\frac{\sqrt{445}}{30} x-\frac{1}{6}=-\frac{\sqrt{445}}{30}$$

Me tāpiri \frac{1}{6} ki ngā taha e rua o te whārite.

$$x=\frac{\sqrt{445}}{30}+\frac{1}{6} x=-\frac{\sqrt{445}}{30}+\frac{1}{6}$$