Whakaoti i te x^2+x+2=5

$${ x }^{ 2 } +x+2 = 5$$

Whakaoti mō x

$$x = \frac{\sqrt{13} - 1}{2} \approx 1.302775638$$

$$x=\frac{-\sqrt{13}-1}{2}\approx -2.302775638$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

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.

$$x^{2}+x+2=5$$

Me tango 5 mai i ngā taha e rua o te whārite.

$$x^{2}+x+2-5=5-5$$

Mā te tango i te 5 i a ia ake anō ka toe ko te 0.

$$x^{2}+x+2-5=0$$

Tango 5 mai i 2.

$$x^{2}+x-3=0$$

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

$$x=\frac{-1±\sqrt{1^{2}-4\left(-3\right)}}{2}$$

Pūrua 1.

$$x=\frac{-1±\sqrt{1-4\left(-3\right)}}{2}$$

Whakareatia -4 ki te -3.

$$x=\frac{-1±\sqrt{1+12}}{2}$$

Tāpiri 1 ki te 12.

$$x=\frac{-1±\sqrt{13}}{2}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{13}}{2} ina he tāpiri te ±. Tāpiri -1 ki te \sqrt{13}.

$$x=\frac{\sqrt{13}-1}{2}$$

Nā, me whakaoti te whārite x=\frac{-1±\sqrt{13}}{2} ina he tango te ±. Tango \sqrt{13} mai i -1.

$$x=\frac{-\sqrt{13}-1}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{13}-1}{2} x=\frac{-\sqrt{13}-1}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

Ko ngā whārite pūrua pēnei i tēnei nā ka taea te whakaoti mā te whakaoti i te pūrua. Hei whakaoti i te pūrua, ko te whārite me mātua tuhi ki te āhua x^{2}+bx=c.

$$x^{2}+x+2=5$$

Me tango 2 mai i ngā taha e rua o te whārite.

$$x^{2}+x+2-2=5-2$$

Mā te tango i te 2 i a ia ake anō ka toe ko te 0.

$$x^{2}+x=5-2$$

Tango 2 mai i 5.

$$x^{2}+x=3$$

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.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=3+\left(\frac{1}{2}\right)^{2}$$

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

$$x^{2}+x+\frac{1}{4}=3+\frac{1}{4}$$

Tāpiri 3 ki te \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{13}{4}$$

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

$$\left(x+\frac{1}{2}\right)^{2}=\frac{13}{4}$$

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

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{13}{4}}$$

Whakarūnātia.

$$x+\frac{1}{2}=\frac{\sqrt{13}}{2} x+\frac{1}{2}=-\frac{\sqrt{13}}{2}$$

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

$$x=\frac{\sqrt{13}-1}{2} x=\frac{-\sqrt{13}-1}{2}$$