Whakaoti i te n(n-1)/2=63

$$\frac { n ( n - 1 ) } { 2 } = 63$$

Whakaoti mō n

$$n = \frac{\sqrt{505} + 1}{2} \approx 11.736102527$$

$$n=\frac{1-\sqrt{505}}{2}\approx -10.736102527$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Me whakarea ngā taha e rua ki te 2.

$$n\left(n-1\right)=63\times 2$$

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

$$n^{2}-n=63\times 2$$

Whakareatia te 63 ki te 2, ka 126.

$$n^{2}-n=126$$

Tangohia te 126 mai i ngā taha e rua.

$$n^{2}-n-126=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -1 mō b, me -126 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-1\right)±\sqrt{1-4\left(-126\right)}}{2}$$

Whakareatia -4 ki te -126.

$$n=\frac{-\left(-1\right)±\sqrt{1+504}}{2}$$

Tāpiri 1 ki te 504.

$$n=\frac{-\left(-1\right)±\sqrt{505}}{2}$$

Ko te tauaro o -1 ko 1.

$$n=\frac{1±\sqrt{505}}{2}$$

Nā, me whakaoti te whārite n=\frac{1±\sqrt{505}}{2} ina he tāpiri te ±. Tāpiri 1 ki te \sqrt{505}.

$$n=\frac{\sqrt{505}+1}{2}$$

Nā, me whakaoti te whārite n=\frac{1±\sqrt{505}}{2} ina he tango te ±. Tango \sqrt{505} mai i 1.

$$n=\frac{1-\sqrt{505}}{2}$$

Kua oti te whārite te whakatau.

$$n=\frac{\sqrt{505}+1}{2} n=\frac{1-\sqrt{505}}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

Me whakarea ngā taha e rua ki te 2.

$$n\left(n-1\right)=63\times 2$$

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

$$n^{2}-n=63\times 2$$

Whakareatia te 63 ki te 2, ka 126.

$$n^{2}-n=126$$

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.

$$n^{2}-n+\left(-\frac{1}{2}\right)^{2}=126+\left(-\frac{1}{2}\right)^{2}$$

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

$$n^{2}-n+\frac{1}{4}=126+\frac{1}{4}$$

Tāpiri 126 ki te \frac{1}{4}.

$$n^{2}-n+\frac{1}{4}=\frac{505}{4}$$

Tauwehea n^{2}-n+\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}.

$$\left(n-\frac{1}{2}\right)^{2}=\frac{505}{4}$$

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

$$\sqrt{\left(n-\frac{1}{2}\right)^{2}}=\sqrt{\frac{505}{4}}$$

Whakarūnātia.

$$n-\frac{1}{2}=\frac{\sqrt{505}}{2} n-\frac{1}{2}=-\frac{\sqrt{505}}{2}$$

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

$$n=\frac{\sqrt{505}+1}{2} n=\frac{1-\sqrt{505}}{2}$$