Whakaoti i te t^2+11+t-42=0

$${ t }^{ 2 } +11+t-42 = 0$$

Whakaoti mō t

$$t = \frac{5 \sqrt{5} - 1}{2} \approx 5.090169944$$

$$t=\frac{-5\sqrt{5}-1}{2}\approx -6.090169944$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 42 i te 11, ka -31.

$$t^{2}-31+t=0$$

Ko ngā whārite katoa o te āhua ax^{2}+bx+c=0 ka taea te whakaoti mā te whakamahi i te tikanga tātai pūrua: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. E rua ngā otinga ka puta i te tikanga tātai pūrua, ko tētahi ina he tāpiri a ±, ā, ko tētahi ina he tango.

$$t^{2}+t-31=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 1 mō b, me -31 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-1±\sqrt{1^{2}-4\left(-31\right)}}{2}$$

Pūrua 1.

$$t=\frac{-1±\sqrt{1-4\left(-31\right)}}{2}$$

Whakareatia -4 ki te -31.

$$t=\frac{-1±\sqrt{1+124}}{2}$$

Tāpiri 1 ki te 124.

$$t=\frac{-1±\sqrt{125}}{2}$$

Tuhia te pūtakerua o te 125.

$$t=\frac{-1±5\sqrt{5}}{2}$$

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

$$t=\frac{5\sqrt{5}-1}{2}$$

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

$$t=\frac{-5\sqrt{5}-1}{2}$$

Kua oti te whārite te whakatau.

$$t=\frac{5\sqrt{5}-1}{2} t=\frac{-5\sqrt{5}-1}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 42 i te 11, ka -31.

$$t^{2}-31+t=0$$

Me tāpiri te 31 ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

$$t^{2}+t=31$$

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.

$$t^{2}+t+\left(\frac{1}{2}\right)^{2}=31+\left(\frac{1}{2}\right)^{2}$$

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

$$t^{2}+t+\frac{1}{4}=31+\frac{1}{4}$$

Tāpiri 31 ki te \frac{1}{4}.

$$t^{2}+t+\frac{1}{4}=\frac{125}{4}$$

Tauwehea t^{2}+t+\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}.

$$\left(t+\frac{1}{2}\right)^{2}=\frac{125}{4}$$

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

$$\sqrt{\left(t+\frac{1}{2}\right)^{2}}=\sqrt{\frac{125}{4}}$$

Whakarūnātia.

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

Me tango \frac{1}{2} mai i ngā taha e rua o te whārite.

$$t=\frac{5\sqrt{5}-1}{2} t=\frac{-5\sqrt{5}-1}{2}$$