Whakaoti i te 0=x^2+11x-8

$$0 = x ^ { 2 } + 11 x - 8$$

Whakaoti mō x

$$x=\frac{3\sqrt{17}-11}{2}\approx 0.684658438$$

$$x=\frac{-3\sqrt{17}-11}{2}\approx -11.684658438$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$x^{2}+11x-8=0$$

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, 11 mō b, me -8 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\left(-8\right)}}{2}$$

Pūrua 11.

$$x=\frac{-11±\sqrt{121-4\left(-8\right)}}{2}$$

Whakareatia -4 ki te -8.

$$x=\frac{-11±\sqrt{121+32}}{2}$$

Tāpiri 121 ki te 32.

$$x=\frac{-11±\sqrt{153}}{2}$$

Tuhia te pūtakerua o te 153.

$$x=\frac{-11±3\sqrt{17}}{2}$$

Nā, me whakaoti te whārite x=\frac{-11±3\sqrt{17}}{2} ina he tāpiri te ±. Tāpiri -11 ki te 3\sqrt{17}.

$$x=\frac{3\sqrt{17}-11}{2}$$

Nā, me whakaoti te whārite x=\frac{-11±3\sqrt{17}}{2} ina he tango te ±. Tango 3\sqrt{17} mai i -11.

$$x=\frac{-3\sqrt{17}-11}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{3\sqrt{17}-11}{2} x=\frac{-3\sqrt{17}-11}{2}$$

Ngā Upane mō te Whakaoti i te Pūrua

Whakawhitihia ngā taha kia puta ki te taha mauī ngā kīanga tau taurangi katoa.

$$x^{2}+11x-8=0$$

Me tāpiri te 8 ki ngā taha e rua. Ko te tau i tāpiria he kore ka hua koia tonu.

$$x^{2}+11x=8$$

Whakawehea te 11, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te \frac{11}{2}. Nā, tāpiria te pūrua o te \frac{11}{2} ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}+11x+\left(\frac{11}{2}\right)^{2}=8+\left(\frac{11}{2}\right)^{2}$$

Pūruatia \frac{11}{2} mā te pūrua i te taurunga me te tauraro o te hautanga.

$$x^{2}+11x+\frac{121}{4}=8+\frac{121}{4}$$

Tāpiri 8 ki te \frac{121}{4}.

$$x^{2}+11x+\frac{121}{4}=\frac{153}{4}$$

Tauwehea te x^{2}+11x+\frac{121}{4}. Ko te tikanga, ina ko x^{2}+bx+c he pūrua tika, ka taea te tauwehe i ngā wā katoa hei \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{11}{2}\right)^{2}=\frac{153}{4}$$

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

$$\sqrt{\left(x+\frac{11}{2}\right)^{2}}=\sqrt{\frac{153}{4}}$$

Whakarūnātia.

$$x+\frac{11}{2}=\frac{3\sqrt{17}}{2} x+\frac{11}{2}=-\frac{3\sqrt{17}}{2}$$

Me tango \frac{11}{2} mai i ngā taha e rua o te whārite.

$$x=\frac{3\sqrt{17}-11}{2} x=\frac{-3\sqrt{17}-11}{2}$$