Question

resolva a seguinte equação biquadrada x^4+ 4abx^2 - ( a^2 - b^2 ) ^2 = 0

Answer 1

Boa tarde Clara

x⁴ + 4abx² - (a² - b²) = 0

y = x²

y² + 4aby - (a² - b²) = 0

delta
d² = 16a²b² + 4a² - 4b²
d = 2*√(4a²b² + a² - b²)

y1 = (-4ab + 2*√(4a²b² + a² - b²)/2
y1 = -2ab + √(4a²b² + a² - b²)

y2 = (-4ab + 2*√(4a²b² + a² - b²)/2
y2 = -2ab -  √(4a²b² + a² - b²)

x1 = √(-2ab + √(4a²b² + a² - b²))
x2 = -√(-2ab + √(4a²b² + a² - b²))
x3 = √(-2ab - √(4a²b² + a² - b²))
x4 = -√(-2ab - √(4a²b² + a² - b²))