x^6+3x^3+2=0 eching

$$x ^ { 6 } + 3 x ^ { 3 } + 2 = 0$$

x uchun yechish (complex solution)

$$x\in \frac{-\sqrt{3}i+1}{2},-1,\frac{1+\sqrt{3}i}{2},\sqrt[3]{2}e^{\frac{5\pi i}{3}},\sqrt[3]{2}e^{\frac{\pi i}{3}},-\sqrt[3]{2}$$

Kvadrat tenglamalaridan foydalanish qoidalari

x^{3} uchun t ni almashtiring.

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

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 3 ni va c uchun 2 ni ayiring.

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

Hisoblarni amalga oshiring.

$$t=\frac{-3±1}{2}$$

t=\frac{-3±1}{2} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$t=-1 t=-2$$

x=t^{3} boʻlganda, yechimlar har bir t uchun tenglamani yechish orqali olinadi.

$$x=-1 x=\frac{1+\sqrt{3}i}{2} x=\frac{-\sqrt{3}i+1}{2} x=-\sqrt[3]{2}ie^{\frac{\pi i}{6}} x=-\sqrt[3]{2} x=\sqrt[3]{2}e^{\frac{\pi i}{3}}$$

x uchun yechish

$$x=-\sqrt[3]{2}\approx -1,25992105$$

$$x=-1$$

Kvadrat tenglamalaridan foydalanish qoidalari

x^{3} uchun t ni almashtiring.

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

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 3 ni va c uchun 2 ni ayiring.

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

Hisoblarni amalga oshiring.

$$t=\frac{-3±1}{2}$$

t=\frac{-3±1}{2} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$t=-1 t=-2$$

x=t^{3} boʻlganda, yechimlar har bir t uchun x=\sqrt[3]{t} hisoblanishi orqali olinadi.

$$x=-1 x=-\sqrt[3]{2}$$