Whakaoti i te x^2+1=4x

$$x ^ { 2 } + 1 = 4 x$$

Whakaoti mō x

$$x=\sqrt{3}+2\approx 3.732050808$$

$$x=2-\sqrt{3}\approx 0.267949192$$

Ngā Upane e Whakamahi ana i te Tikanga Tātai Pūrua

Tangohia te 4x mai i ngā taha e rua.

$$x^{2}+1-4x=0$$

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.

$$x^{2}-4x+1=0$$

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

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

Pūrua -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4}}{2}$$

Tāpiri 16 ki te -4.

$$x=\frac{-\left(-4\right)±\sqrt{12}}{2}$$

Tuhia te pūtakerua o te 12.

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

Ko te tauaro o -4 ko 4.

$$x=\frac{4±2\sqrt{3}}{2}$$

Nā, me whakaoti te whārite x=\frac{4±2\sqrt{3}}{2} ina he tāpiri te ±. Tāpiri 4 ki te 2\sqrt{3}.

$$x=\frac{2\sqrt{3}+4}{2}$$

Whakawehe 4+2\sqrt{3} ki te 2.

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

Nā, me whakaoti te whārite x=\frac{4±2\sqrt{3}}{2} ina he tango te ±. Tango 2\sqrt{3} mai i 4.

$$x=\frac{4-2\sqrt{3}}{2}$$

Whakawehe 4-2\sqrt{3} ki te 2.

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

Kua oti te whārite te whakatau.

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

Ngā Upane mō te Whakaoti i te Pūrua

Tangohia te 4x mai i ngā taha e rua.

$$x^{2}+1-4x=0$$

Tangohia te 1 mai i ngā taha e rua. Ko te tau i tango i te kore ka hua ko tōna korenga.

$$x^{2}-4x=-1$$

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

$$x^{2}-4x+\left(-2\right)^{2}=-1+\left(-2\right)^{2}$$

Pūrua -2.

$$x^{2}-4x+4=-1+4$$

Tāpiri -1 ki te 4.

$$x^{2}-4x+4=3$$

Tauwehea te x^{2}-4x+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}.

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

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

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{3}$$

Whakarūnātia.

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

Me tāpiri 2 ki ngā taha e rua o te whārite.

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