Whakaoti i te x^2+5x+1=0

$${ x }^{ 2 } +5x+1=0$$

Whakaoti mō x

$$x=\frac{\sqrt{21}-5}{2}\approx -0.208712153$$

$$x=\frac{-\sqrt{21}-5}{2}\approx -4.791287847$$

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.

$$x^{2}+5x+1=0$$

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

$$x=\frac{-5±\sqrt{5^{2}-4}}{2}$$

Pūrua 5.

$$x=\frac{-5±\sqrt{25-4}}{2}$$

Tāpiri 25 ki te -4.

$$x=\frac{-5±\sqrt{21}}{2}$$

Nā, me whakaoti te whārite x=\frac{-5±\sqrt{21}}{2} ina he tāpiri te ±. Tāpiri -5 ki te \sqrt{21}.

$$x=\frac{\sqrt{21}-5}{2}$$

Nā, me whakaoti te whārite x=\frac{-5±\sqrt{21}}{2} ina he tango te ±. Tango \sqrt{21} mai i -5.

$$x=\frac{-\sqrt{21}-5}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{\sqrt{21}-5}{2} x=\frac{-\sqrt{21}-5}{2}$$

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.

$$x^{2}+5x+1=0$$

Me tango 1 mai i ngā taha e rua o te whārite.

$$x^{2}+5x+1-1=-1$$

Mā te tango i te 1 i a ia ake anō ka toe ko te 0.

$$x^{2}+5x=-1$$

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.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=-1+\left(\frac{5}{2}\right)^{2}$$

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

$$x^{2}+5x+\frac{25}{4}=-1+\frac{25}{4}$$

Tāpiri -1 ki te \frac{25}{4}.

$$x^{2}+5x+\frac{25}{4}=\frac{21}{4}$$

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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{21}{4}$$

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

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{21}{4}}$$

Whakarūnātia.

$$x+\frac{5}{2}=\frac{\sqrt{21}}{2} x+\frac{5}{2}=-\frac{\sqrt{21}}{2}$$

Me tango \frac{5}{2} mai i ngā taha e rua o te whārite.

$$x=\frac{\sqrt{21}-5}{2} x=\frac{-\sqrt{21}-5}{2}$$