Whakaoti i te -48=n/2(2(9)(n-1)-2)

$$- 48 = \frac { n } { 2 } ( 2 ( 9 ) ( n - 1 ) - 2 )$$

Whakaoti mō n

$$n=\frac{5+\sqrt{407}i}{9}\approx 0.555555556+2.241582334i$$

$$n=\frac{-\sqrt{407}i+5}{9}\approx 0.555555556-2.241582334i$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakareatia ngā taha e rua o te whārite ki te 2.

$$-96=n\left(2\times 9\left(n-1\right)-2\right)$$

Whakareatia te 2 ki te 9, ka 18.

$$-96=n\left(18\left(n-1\right)-2\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 18 ki te n-1.

$$-96=n\left(18n-18-2\right)$$

Tangohia te 2 i te -18, ka -20.

$$-96=n\left(18n-20\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te n ki te 18n-20.

$$-96=18n^{2}-20n$$

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

$$18n^{2}-20n=-96$$

Me tāpiri te 96 ki ngā taha e rua.

$$18n^{2}-20n+96=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 18 mō a, -20 mō b, me 96 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 18\times 96}}{2\times 18}$$

Pūrua -20.

$$n=\frac{-\left(-20\right)±\sqrt{400-4\times 18\times 96}}{2\times 18}$$

Whakareatia -4 ki te 18.

$$n=\frac{-\left(-20\right)±\sqrt{400-72\times 96}}{2\times 18}$$

Whakareatia -72 ki te 96.

$$n=\frac{-\left(-20\right)±\sqrt{400-6912}}{2\times 18}$$

Tāpiri 400 ki te -6912.

$$n=\frac{-\left(-20\right)±\sqrt{-6512}}{2\times 18}$$

Tuhia te pūtakerua o te -6512.

$$n=\frac{-\left(-20\right)±4\sqrt{407}i}{2\times 18}$$

Ko te tauaro o -20 ko 20.

$$n=\frac{20±4\sqrt{407}i}{2\times 18}$$

Whakareatia 2 ki te 18.

$$n=\frac{20±4\sqrt{407}i}{36}$$

Nā, me whakaoti te whārite n=\frac{20±4\sqrt{407}i}{36} ina he tāpiri te ±. Tāpiri 20 ki te 4i\sqrt{407}.

$$n=\frac{20+4\sqrt{407}i}{36}$$

Whakawehe 20+4i\sqrt{407} ki te 36.

$$n=\frac{5+\sqrt{407}i}{9}$$

Nā, me whakaoti te whārite n=\frac{20±4\sqrt{407}i}{36} ina he tango te ±. Tango 4i\sqrt{407} mai i 20.

$$n=\frac{-4\sqrt{407}i+20}{36}$$

Whakawehe 20-4i\sqrt{407} ki te 36.

$$n=\frac{-\sqrt{407}i+5}{9}$$

Kua oti te whārite te whakatau.

$$n=\frac{5+\sqrt{407}i}{9} n=\frac{-\sqrt{407}i+5}{9}$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakareatia ngā taha e rua o te whārite ki te 2.

$$-96=n\left(2\times 9\left(n-1\right)-2\right)$$

Whakareatia te 2 ki te 9, ka 18.

$$-96=n\left(18\left(n-1\right)-2\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te 18 ki te n-1.

$$-96=n\left(18n-18-2\right)$$

Tangohia te 2 i te -18, ka -20.

$$-96=n\left(18n-20\right)$$

Whakamahia te āhuatanga tohatoha hei whakarea te n ki te 18n-20.

$$-96=18n^{2}-20n$$

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

$$18n^{2}-20n=-96$$

Whakawehea ngā taha e rua ki te 18.

$$\frac{18n^{2}-20n}{18}=-\frac{96}{18}$$

Mā te whakawehe ki te 18 ka wetekia te whakareanga ki te 18.

$$n^{2}+\left(-\frac{20}{18}\right)n=-\frac{96}{18}$$

Whakahekea te hautanga \frac{-20}{18} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 2.

$$n^{2}-\frac{10}{9}n=-\frac{96}{18}$$

Whakahekea te hautanga \frac{-96}{18} ki ōna wāhi pāpaku rawa mā te tango me te whakakore i te 6.

$$n^{2}-\frac{10}{9}n=-\frac{16}{3}$$

Whakawehea te -\frac{10}{9}, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -\frac{5}{9}. Nā, tāpiria te pūrua o te -\frac{5}{9} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$n^{2}-\frac{10}{9}n+\left(-\frac{5}{9}\right)^{2}=-\frac{16}{3}+\left(-\frac{5}{9}\right)^{2}$$

Pūruatia -\frac{5}{9} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$n^{2}-\frac{10}{9}n+\frac{25}{81}=-\frac{16}{3}+\frac{25}{81}$$

Tāpiri -\frac{16}{3} ki te \frac{25}{81} 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.

$$n^{2}-\frac{10}{9}n+\frac{25}{81}=-\frac{407}{81}$$

Tauwehea n^{2}-\frac{10}{9}n+\frac{25}{81}. 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}.

$$\left(n-\frac{5}{9}\right)^{2}=-\frac{407}{81}$$

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

$$\sqrt{\left(n-\frac{5}{9}\right)^{2}}=\sqrt{-\frac{407}{81}}$$

Whakarūnātia.

$$n-\frac{5}{9}=\frac{\sqrt{407}i}{9} n-\frac{5}{9}=-\frac{\sqrt{407}i}{9}$$

Me tāpiri \frac{5}{9} ki ngā taha e rua o te whārite.

$$n=\frac{5+\sqrt{407}i}{9} n=\frac{-\sqrt{407}i+5}{9}$$