4(x^4+1)=5x^2 eching

$$4 ( x ^ { 4 } + 1 ) = 5 x ^ { 2 }$$

x uchun yechish (complex solution)

$$x=e^{\frac{-\arctan(\frac{\sqrt{39}}{5})i+2\pi i}{2}}\approx -0,901387819+0,433012702i$$

$$x=e^{-\frac{\arctan(\frac{\sqrt{39}}{5})i}{2}}\approx 0,901387819-0,433012702i$$

$$x=e^{\frac{\arctan(\frac{\sqrt{39}}{5})i+2\pi i}{2}}\approx -0,901387819-0,433012702i$$

$$x=e^{\frac{\arctan(\frac{\sqrt{39}}{5})i}{2}}\approx 0,901387819+0,433012702i$$

Kvadrat tenglamalaridan foydalanish qoidalari

4 ga x^{4}+1 ni ko'paytirish orqali distributiv xususiyatdan foydalanish.

$$4x^{4}+4=5x^{2}$$

Ikkala tarafdan 5x^{2} ni ayirish.

$$4x^{4}+4-5x^{2}=0$$

x^{2} uchun t ni almashtiring.

$$4t^{2}-5t+4=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni bu formula bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat tenglamada a uchun 4 ni, b uchun -5 ni va c uchun 4 ni ayiring.

$$t=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 4\times 4}}{2\times 4}$$

Hisoblarni amalga oshiring.

$$t=\frac{5±\sqrt{-39}}{8}$$

t=\frac{5±\sqrt{-39}}{8} tenglamasini ± plus va ± minus boʻlgan holatida ishlang.

$$t=\frac{5+\sqrt{39}i}{8} t=\frac{-\sqrt{39}i+5}{8}$$

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

$$x=e^{\frac{\arctan(\frac{\sqrt{39}}{5})i+2\pi i}{2}} x=e^{\frac{\arctan(\frac{\sqrt{39}}{5})i}{2}} x=e^{-\frac{\arctan(\frac{\sqrt{39}}{5})i}{2}} x=e^{\frac{-\arctan(\frac{\sqrt{39}}{5})i+2\pi i}{2}}$$