z^4-2z^2+3=0 eching

$${ z }^{ 4 } -2 { z }^{ 2 } +3=0$$

z uchun yechish

$$z=\sqrt[4]{3}e^{\frac{-\arctan(\sqrt{2})i+2\pi i}{2}}\approx -1,168770894+0,605000334i$$

$$z=\sqrt[4]{3}e^{-\frac{\arctan(\sqrt{2})i}{2}}\approx 1,168770894-0,605000334i$$

$$z=\sqrt[4]{3}e^{\frac{\arctan(\sqrt{2})i+2\pi i}{2}}\approx -1,168770894-0,605000334i$$

$$z=\sqrt[4]{3}e^{\frac{\arctan(\sqrt{2})i}{2}}\approx 1,168770894+0,605000334i$$

Kvadrat tenglamalaridan foydalanish qoidalari

z^{2} uchun t ni almashtiring.

$$t^{2}-2t+3=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun 1 ni, b uchun -2 ni va c uchun 3 ni ayiring.

$$t=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 1\times 3}}{2}$$

Hisoblarni amalga oshiring.

$$t=\frac{2±\sqrt{-8}}{2}$$

t=\frac{2±\sqrt{-8}}{2} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$t=1+\sqrt{2}i t=-\sqrt{2}i+1$$

z=t^{2} boʻlganda, yechimlar har bir t uchun z=±\sqrt{t} hisoblanishi orqali olinadi.

$$z=\sqrt[4]{3}e^{\frac{\arctan(\sqrt{2})i+2\pi i}{2}} z=\sqrt[4]{3}e^{\frac{\arctan(\sqrt{2})i}{2}} z=\sqrt[4]{3}e^{-\frac{\arctan(\sqrt{2})i}{2}} z=\sqrt[4]{3}e^{\frac{-\arctan(\sqrt{2})i+2\pi i}{2}}$$