Question

3x4 -2x2-16=0..................

Answer 1

Решение
3x⁴  - 2x² - 16 = 0
Пусть x² = y, тогда
3y² - 2y - 16 = 0
D = 2² + 4*3*16 = 196
y₁ = (2 - 14)/6
y₁ = - 2
y₂ = (2 + 14)/6
y₂ = 16/6
y₂ = 2(2/3)
x² = - 2 не имеет смысла
x² = (8/3)
x₁ = -√(8/3)
x₁ = - 2√(2/3)
x₁ = - 2√6 / 3
x₂ = √(8/3)
x₂ = 2√(2/3)
x₂ = 2√6 / 3

Answer 2

To solve the quadratic equation 3x^4 - 2x^2 - 16 = 0, we can first substitute a variable to make it easier to solve. Let's use a substitution: Let y = x^2.

Now, the equation becomes 3y^2 - 2y - 16 = 0.

To solve this quadratic equation, we can use the quadratic formula:

y = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 3, b = -2, and c = -16. Plugging in these values, we get:

y = (-(-2) ± √((-2)^2 - 4 * 3 * (-16))) / (2 * 3) = (2 ± √(4 + 192)) / 6 = (2 ± √196) / 6 = (2 ± 14) / 6

Simplifying further, we have two possible solutions for y:

1. y = (2 + 14) / 6 = 16 / 6 = 8/3
2. y = (2 - 14) / 6 = -12 / 6 = -2

Now, we substitute y back in terms of x:

1. If y = 8/3: x^2 = 8/3 x = ±√(8/3) = ±(2√2) / √3 = ±(2√2/√3) = ±(2√6) / 3

2. If y = -2: x^2 = -2 x = ±√(-2) = ±i√2, where i is the imaginary unit.

Therefore, the solutions to the equation 3x^4 - 2x^2 - 16 = 0 are: x = (2√6) / 3, -(2√6) / 3, i√2, -i√2