Whakaoti i te x^2-5x-7=0

$$x ^ { 2 } - 5 x - 7 = 0$$

Whakaoti mō x

$$x = \frac{\sqrt{53} + 5}{2} \approx 6.140054945$$

$$x=\frac{5-\sqrt{53}}{2}\approx -1.140054945$$

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.

$$x^{2}-5x-7=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -5 mō b, me -7 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-7\right)}}{2}$$

Pūrua -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-7\right)}}{2}$$

Whakareatia -4 ki te -7.

$$x=\frac{-\left(-5\right)±\sqrt{25+28}}{2}$$

Tāpiri 25 ki te 28.

$$x=\frac{-\left(-5\right)±\sqrt{53}}{2}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±\sqrt{53}}{2}$$

Nā, me whakaoti te whārite x=\frac{5±\sqrt{53}}{2} ina he tāpiri te ±. Tāpiri 5 ki te \sqrt{53}.

$$x=\frac{\sqrt{53}+5}{2}$$

Nā, me whakaoti te whārite x=\frac{5±\sqrt{53}}{2} ina he tango te ±. Tango \sqrt{53} mai i 5.

$$x=\frac{5-\sqrt{53}}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{53}+5}{2} x=\frac{5-\sqrt{53}}{2}$$

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.

$$x^{2}-5x-7=0$$

Me tāpiri 7 ki ngā taha e rua o te whārite.

$$x^{2}-5x-7-\left(-7\right)=-\left(-7\right)$$

Mā te tango i te -7 i a ia ake anō ka toe ko te 0.

$$x^{2}-5x=-\left(-7\right)$$

Tango -7 mai i 0.

$$x^{2}-5x=7$$

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.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=7+\left(-\frac{5}{2}\right)^{2}$$

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

$$x^{2}-5x+\frac{25}{4}=7+\frac{25}{4}$$

Tāpiri 7 ki te \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{53}{4}$$

Tauwehea x^{2}-5x+\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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{53}{4}$$

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

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{53}{4}}$$

Whakarūnātia.

$$x-\frac{5}{2}=\frac{\sqrt{53}}{2} x-\frac{5}{2}=-\frac{\sqrt{53}}{2}$$

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

$$x=\frac{\sqrt{53}+5}{2} x=\frac{5-\sqrt{53}}{2}$$