Whakaoti i te x^2-x/9=2/15

$$\frac { x ^ { 2 } - x } { 9 } = \frac { 2 } { 15 }$$

Whakaoti mō x

$$x=\frac{\sqrt{145}}{10}+\frac{1}{2}\approx 1.704159458$$

$$x=-\frac{\sqrt{145}}{10}+\frac{1}{2}\approx -0.704159458$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Me whakarea ngā taha e rua ki te 9.

$$x^{2}-x=\frac{2}{15}\times 9$$

Whakareatia te \frac{2}{15} ki te 9, ka \frac{6}{5}.

$$x^{2}-x=\frac{6}{5}$$

Tangohia te \frac{6}{5} mai i ngā taha e rua.

$$x^{2}-x-\frac{6}{5}=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -1 mō b, me -\frac{6}{5} mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\left(-\frac{6}{5}\right)}}{2}$$

Whakareatia -4 ki te -\frac{6}{5}.

$$x=\frac{-\left(-1\right)±\sqrt{1+\frac{24}{5}}}{2}$$

Tāpiri 1 ki te \frac{24}{5}.

$$x=\frac{-\left(-1\right)±\sqrt{\frac{29}{5}}}{2}$$

Tuhia te pūtakerua o te \frac{29}{5}.

$$x=\frac{-\left(-1\right)±\frac{\sqrt{145}}{5}}{2}$$

Ko te tauaro o -1 ko 1.

$$x=\frac{1±\frac{\sqrt{145}}{5}}{2}$$

Nā, me whakaoti te whārite x=\frac{1±\frac{\sqrt{145}}{5}}{2} ina he tāpiri te ±. Tāpiri 1 ki te \frac{\sqrt{145}}{5}.

$$x=\frac{\frac{\sqrt{145}}{5}+1}{2}$$

Whakawehe 1+\frac{\sqrt{145}}{5} ki te 2.

$$x=\frac{\sqrt{145}}{10}+\frac{1}{2}$$

Nā, me whakaoti te whārite x=\frac{1±\frac{\sqrt{145}}{5}}{2} ina he tango te ±. Tango \frac{\sqrt{145}}{5} mai i 1.

$$x=\frac{-\frac{\sqrt{145}}{5}+1}{2}$$

Whakawehe 1-\frac{\sqrt{145}}{5} ki te 2.

$$x=-\frac{\sqrt{145}}{10}+\frac{1}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{145}}{10}+\frac{1}{2} x=-\frac{\sqrt{145}}{10}+\frac{1}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

Me whakarea ngā taha e rua ki te 9.

$$x^{2}-x=\frac{2}{15}\times 9$$

Whakareatia te \frac{2}{15} ki te 9, ka \frac{6}{5}.

$$x^{2}-x=\frac{6}{5}$$

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.

$$x^{2}-x+\left(-\frac{1}{2}\right)^{2}=\frac{6}{5}+\left(-\frac{1}{2}\right)^{2}$$

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

$$x^{2}-x+\frac{1}{4}=\frac{6}{5}+\frac{1}{4}$$

Tāpiri \frac{6}{5} 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.

$$x^{2}-x+\frac{1}{4}=\frac{29}{20}$$

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}.

$$\left(x-\frac{1}{2}\right)^{2}=\frac{29}{20}$$

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

$$\sqrt{\left(x-\frac{1}{2}\right)^{2}}=\sqrt{\frac{29}{20}}$$

Whakarūnātia.

$$x-\frac{1}{2}=\frac{\sqrt{145}}{10} x-\frac{1}{2}=-\frac{\sqrt{145}}{10}$$

Me tāpiri \frac{1}{2} ki ngā taha e rua o te whārite.

$$x=\frac{\sqrt{145}}{10}+\frac{1}{2} x=-\frac{\sqrt{145}}{10}+\frac{1}{2}$$