Whakaoti i te (n+2)(n+1)=145

$$( n + 2 ) ( n + 1 ) = 145$$

Whakaoti mō n

$$n = \frac{\sqrt{581} - 3}{2} \approx 10.551970793$$

$$n=\frac{-\sqrt{581}-3}{2}\approx -13.551970793$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakamahia te āhuatanga tuaritanga hei whakarea te n+2 ki te n+1 ka whakakotahi i ngā kupu rite.

$$n^{2}+3n+2=145$$

Tangohia te 145 mai i ngā taha e rua.

$$n^{2}+3n+2-145=0$$

Tangohia te 145 i te 2, ka -143.

$$n^{2}+3n-143=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 3 mō b, me -143 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-3±\sqrt{3^{2}-4\left(-143\right)}}{2}$$

Pūrua 3.

$$n=\frac{-3±\sqrt{9-4\left(-143\right)}}{2}$$

Whakareatia -4 ki te -143.

$$n=\frac{-3±\sqrt{9+572}}{2}$$

Tāpiri 9 ki te 572.

$$n=\frac{-3±\sqrt{581}}{2}$$

Nā, me whakaoti te whārite n=\frac{-3±\sqrt{581}}{2} ina he tāpiri te ±. Tāpiri -3 ki te \sqrt{581}.

$$n=\frac{\sqrt{581}-3}{2}$$

Nā, me whakaoti te whārite n=\frac{-3±\sqrt{581}}{2} ina he tango te ±. Tango \sqrt{581} mai i -3.

$$n=\frac{-\sqrt{581}-3}{2}$$

Kua oti te whārite te whakatau.

$$n=\frac{\sqrt{581}-3}{2} n=\frac{-\sqrt{581}-3}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakamahia te āhuatanga tuaritanga hei whakarea te n+2 ki te n+1 ka whakakotahi i ngā kupu rite.

$$n^{2}+3n+2=145$$

Tangohia te 2 mai i ngā taha e rua.

$$n^{2}+3n=145-2$$

Tangohia te 2 i te 145, ka 143.

$$n^{2}+3n=143$$

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

$$n^{2}+3n+\left(\frac{3}{2}\right)^{2}=143+\left(\frac{3}{2}\right)^{2}$$

Pūruatia \frac{3}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$n^{2}+3n+\frac{9}{4}=143+\frac{9}{4}$$

Tāpiri 143 ki te \frac{9}{4}.

$$n^{2}+3n+\frac{9}{4}=\frac{581}{4}$$

Tauwehea n^{2}+3n+\frac{9}{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}.

$$\left(n+\frac{3}{2}\right)^{2}=\frac{581}{4}$$

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

$$\sqrt{\left(n+\frac{3}{2}\right)^{2}}=\sqrt{\frac{581}{4}}$$

Whakarūnātia.

$$n+\frac{3}{2}=\frac{\sqrt{581}}{2} n+\frac{3}{2}=-\frac{\sqrt{581}}{2}$$

Me tango \frac{3}{2} mai i ngā taha e rua o te whārite.

$$n=\frac{\sqrt{581}-3}{2} n=\frac{-\sqrt{581}-3}{2}$$