32x^10-97x^5+3=0 eching

$$32 x ^ { 10 } - 97 x ^ { 5 } + 3 = 0$$

x uchun yechish (complex solution)

$$x\in \sqrt[5]{3},\sqrt[5]{3}e^{\frac{8\pi i}{5}},\sqrt[5]{3}e^{\frac{2\pi i}{5}},\sqrt[5]{3}e^{\frac{4\pi i}{5}},\sqrt[5]{3}e^{\frac{6\pi i}{5}},\frac{e^{\frac{8\pi i}{5}}}{2},\frac{1}{2},\frac{e^{\frac{2\pi i}{5}}}{2},\frac{e^{\frac{4\pi i}{5}}}{2},\frac{e^{\frac{6\pi i}{5}}}{2}$$

Kvadrat tenglamalaridan foydalanish qoidalari

x^{5} uchun t ni almashtiring.

$$32t^{2}-97t+3=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun 32 ni, b uchun -97 ni va c uchun 3 ni ayiring.

$$t=\frac{-\left(-97\right)±\sqrt{\left(-97\right)^{2}-4\times 32\times 3}}{2\times 32}$$

Hisoblarni amalga oshiring.

$$t=\frac{97±95}{64}$$

t=\frac{97±95}{64} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$t=3 t=\frac{1}{32}$$

x=t^{5} boʻlganda, yechimlar har bir t uchun tenglamani yechish orqali olinadi.

$$x=-\sqrt[5]{3}ie^{\frac{\pi i}{10}} x=-\sqrt[5]{3}e^{\frac{\pi i}{5}} x=\sqrt[5]{3}ie^{\frac{3\pi i}{10}} x=\sqrt[5]{3}e^{\frac{2\pi i}{5}} x=\sqrt[5]{3} x=-\frac{ie^{\frac{\pi i}{10}}}{2} x=-\frac{e^{\frac{\pi i}{5}}}{2} x=\frac{ie^{\frac{3\pi i}{10}}}{2} x=\frac{e^{\frac{2\pi i}{5}}}{2} x=\frac{1}{2}$$

x uchun yechish

$$x=\frac{1}{2}=0,5$$

$$x=\sqrt[5]{3}\approx 1,24573094$$

Kvadrat tenglamalaridan foydalanish qoidalari

x^{5} uchun t ni almashtiring.

$$32t^{2}-97t+3=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun 32 ni, b uchun -97 ni va c uchun 3 ni ayiring.

$$t=\frac{-\left(-97\right)±\sqrt{\left(-97\right)^{2}-4\times 32\times 3}}{2\times 32}$$

Hisoblarni amalga oshiring.

$$t=\frac{97±95}{64}$$

t=\frac{97±95}{64} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$t=3 t=\frac{1}{32}$$

x=t^{5} boʻlganda, yechimlar har bir t uchun x=\sqrt[5]{t} hisoblanishi orqali olinadi.

$$x=\sqrt[5]{3} x=\frac{1}{2}$$