y^4+16y^2-15=0
(y^2 + 8)^2 = 79
y^2 = -8 ± √79
y = ± √(-8 ± √79)
y⁴ +16 y² -15 = 0
( y² )² +16 y² -15 = 0
Substitute y² = x
x² +16 x -15 = 0
x² +16 x = 15
x² +16 x + 64 = 15 + 64
x² +16 x + 64 = 79
( x + 8 )² = 79
x + 8 = ± √79
x = ± √79 - 8
x₁= - √79 - 8
x₂= √79 - 8
Now:
y² = x
y² = x₁
y² = - √79 - 8
y = ± √( - √79 - 8 )
y = ± √ [ ( - 1 ) ∙ √( √79 + 8 ) ]
y = ± √( - 1 ) ∙ √( √79 + 8 )
y = ± i ∙ √( √79 + 8 )
y₁ = - i ∙ √( √79 + 8 )
y₂ = i ∙ √( √79 + 8 )
y² = x
y² = x₂
y² = √79 - 8
y = ± √(√79 - 8 )
y₃ = - √(√79 - 8 )
y₄ = √( √79 - 8 )
The solutions are:
- i ∙ √( √79 + 8 ) , i ∙ √( √79 + 8 ) , - √( √79 - 8 ) , √( √79 - 8 )
To solve the equation y^4 + 16y^2 - 15 = 0, we can use a substitution. Let z = y^2.
Substituting z into the equation, we get:
z^2 + 16z - 15 = 0.
Now we have a quadratic equation in terms of z. To solve this quadratic equation, we can use factoring, completing the square, or the quadratic formula.
Factoring:
The equation z^2 + 16z - 15 = 0 can be factored as (z + 5)(z - 3) = 0.
Setting each factor equal to zero gives us z + 5 = 0 or z - 3 = 0.
Solving for z, we find z = -5 or z = 3.
Now we substitute back y^2 for z to get the solutions for y:
For z = -5:
y^2 = -5
Since the square of a real number cannot be negative, there are no solutions for this case.
For z = 3:
y^2 = 3
Taking the square root of both sides, we obtain:
y = ±sqrt(3).
Therefore, the solutions for the equation y^4 + 16y^2 - 15 = 0 are y = sqrt(3) and y = -sqrt(3).