Whakaoti i te 5(t^2-5t-16)=0

$$5 ( t ^ { 2 } - 5 t - 16 ) = 0$$

Whakaoti mō t

$$t = \frac{\sqrt{89} + 5}{2} \approx 7.216990566$$

$$t=\frac{5-\sqrt{89}}{2}\approx -2.216990566$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakawehea ngā taha e rua ki te 5. Ko te kore i whakawehea ki te tau ehara te kore ka hua ko te kore.

$$t^{2}-5t-16=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -5 mō b, me -16 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-16\right)}}{2}$$

Pūrua -5.

$$t=\frac{-\left(-5\right)±\sqrt{25-4\left(-16\right)}}{2}$$

Whakareatia -4 ki te -16.

$$t=\frac{-\left(-5\right)±\sqrt{25+64}}{2}$$

Tāpiri 25 ki te 64.

$$t=\frac{-\left(-5\right)±\sqrt{89}}{2}$$

Ko te tauaro o -5 ko 5.

$$t=\frac{5±\sqrt{89}}{2}$$

Nā, me whakaoti te whārite t=\frac{5±\sqrt{89}}{2} ina he tāpiri te ±. Tāpiri 5 ki te \sqrt{89}.

$$t=\frac{\sqrt{89}+5}{2}$$

Nā, me whakaoti te whārite t=\frac{5±\sqrt{89}}{2} ina he tango te ±. Tango \sqrt{89} mai i 5.

$$t=\frac{5-\sqrt{89}}{2}$$

Kua oti te whārite te whakatau.

$$t=\frac{\sqrt{89}+5}{2} t=\frac{5-\sqrt{89}}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakawehea ngā taha e rua ki te 5. Ko te kore i whakawehea ki te tau ehara te kore ka hua ko te kore.

$$t^{2}-5t-16=0$$

Me tāpiri te 16 ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

$$t^{2}-5t=16$$

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

$$t^{2}-5t+\left(-\frac{5}{2}\right)^{2}=16+\left(-\frac{5}{2}\right)^{2}$$

Pūruatia -\frac{5}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$t^{2}-5t+\frac{25}{4}=16+\frac{25}{4}$$

Tāpiri 16 ki te \frac{25}{4}.

$$t^{2}-5t+\frac{25}{4}=\frac{89}{4}$$

Tauwehea t^{2}-5t+\frac{25}{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}.

$$\left(t-\frac{5}{2}\right)^{2}=\frac{89}{4}$$

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

$$\sqrt{\left(t-\frac{5}{2}\right)^{2}}=\sqrt{\frac{89}{4}}$$

Whakarūnātia.

$$t-\frac{5}{2}=\frac{\sqrt{89}}{2} t-\frac{5}{2}=-\frac{\sqrt{89}}{2}$$

Me tāpiri \frac{5}{2} ki ngā taha e rua o te whārite.

$$t=\frac{\sqrt{89}+5}{2} t=\frac{5-\sqrt{89}}{2}$$