Whakaoti i te x^2+3=8x

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

Whakaoti mō x

$$x=\sqrt{13}+4\approx 7.605551275$$

$$x=4-\sqrt{13}\approx 0.394448725$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 8x mai i ngā taha e rua.

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

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.

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

Kei te āhua arowhānui tēnei whārite: ax^{2}+bx+c=0. Me whakakapi 1 mō a, -8 mō b, me 3 mō c i te tikanga tātai pūrua, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 3}}{2}$$

Pūrua -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 3}}{2}$$

Whakareatia -4 ki te 3.

$$x=\frac{-\left(-8\right)±\sqrt{64-12}}{2}$$

Tāpiri 64 ki te -12.

$$x=\frac{-\left(-8\right)±\sqrt{52}}{2}$$

Tuhia te pūtakerua o te 52.

$$x=\frac{-\left(-8\right)±2\sqrt{13}}{2}$$

Ko te tauaro o -8 ko 8.

$$x=\frac{8±2\sqrt{13}}{2}$$

Nā, me whakaoti te whārite x=\frac{8±2\sqrt{13}}{2} ina he tāpiri te ±. Tāpiri 8 ki te 2\sqrt{13}.

$$x=\frac{2\sqrt{13}+8}{2}$$

Whakawehe 8+2\sqrt{13} ki te 2.

$$x=\sqrt{13}+4$$

Nā, me whakaoti te whārite x=\frac{8±2\sqrt{13}}{2} ina he tango te ±. Tango 2\sqrt{13} mai i 8.

$$x=\frac{8-2\sqrt{13}}{2}$$

Whakawehe 8-2\sqrt{13} ki te 2.

$$x=4-\sqrt{13}$$

Kua oti te whārite te whakatau.

$$x=\sqrt{13}+4 x=4-\sqrt{13}$$

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 8x mai i ngā taha e rua.

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

Tangohia te 3 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

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

Whakawehea te -8, te tau whakarea o te kīanga tau x, ki te 2 kia riro ai te -4. Nā, tāpiria te pūrua o te -4 ki ngā taha e rua o te whārite. Mā konei e pūrua tika tonu ai te taha mauī o te whārite.

$$x^{2}-8x+\left(-4\right)^{2}=-3+\left(-4\right)^{2}$$

Pūrua -4.

$$x^{2}-8x+16=-3+16$$

Tāpiri -3 ki te 16.

$$x^{2}-8x+16=13$$

Tauwehea x^{2}-8x+16. 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}.

$$\left(x-4\right)^{2}=13$$

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

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{13}$$

Whakarūnātia.

$$x-4=\sqrt{13} x-4=-\sqrt{13}$$

Me tāpiri 4 ki ngā taha e rua o te whārite.

$$x=\sqrt{13}+4 x=4-\sqrt{13}$$