Whakaoti i te (x^2-5x+3)=8

$$( { x }^{ 2 } -5x+3)=8$$

Whakaoti mō x

$$x = \frac{3 \sqrt{5} + 5}{2} \approx 5.854101966$$

$$x=\frac{5-3\sqrt{5}}{2}\approx -0.854101966$$

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.

$$x^{2}-5x+3=8$$

Me tango 8 mai i ngā taha e rua o te whārite.

$$x^{2}-5x+3-8=8-8$$

Mā te tango i te 8 i a ia ake anō ka toe ko te 0.

$$x^{2}-5x+3-8=0$$

Tango 8 mai i 3.

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

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

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

Pūrua -5.

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

Whakareatia -4 ki te -5.

$$x=\frac{-\left(-5\right)±\sqrt{25+20}}{2}$$

Tāpiri 25 ki te 20.

$$x=\frac{-\left(-5\right)±\sqrt{45}}{2}$$

Tuhia te pūtakerua o te 45.

$$x=\frac{-\left(-5\right)±3\sqrt{5}}{2}$$

Ko te tauaro o -5 ko 5.

$$x=\frac{5±3\sqrt{5}}{2}$$

Nā, me whakaoti te whārite x=\frac{5±3\sqrt{5}}{2} ina he tāpiri te ±. Tāpiri 5 ki te 3\sqrt{5}.

$$x=\frac{3\sqrt{5}+5}{2}$$

Nā, me whakaoti te whārite x=\frac{5±3\sqrt{5}}{2} ina he tango te ±. Tango 3\sqrt{5} mai i 5.

$$x=\frac{5-3\sqrt{5}}{2}$$

Kua oti te whārite te whakatau.

$$x=\frac{3\sqrt{5}+5}{2} x=\frac{5-3\sqrt{5}}{2}$$

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.

$$x^{2}-5x+3=8$$

Me tango 3 mai i ngā taha e rua o te whārite.

$$x^{2}-5x+3-3=8-3$$

Mā te tango i te 3 i a ia ake anō ka toe ko te 0.

$$x^{2}-5x=8-3$$

Tango 3 mai i 8.

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

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.

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

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

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

Tāpiri 5 ki te \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{45}{4}$$

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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{45}{4}$$

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

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{45}{4}}$$

Whakarūnātia.

$$x-\frac{5}{2}=\frac{3\sqrt{5}}{2} x-\frac{5}{2}=-\frac{3\sqrt{5}}{2}$$

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

$$x=\frac{3\sqrt{5}+5}{2} x=\frac{5-3\sqrt{5}}{2}$$