9x^-4-35x^-2-4=0
I assume we are solving?
multiply each term by x^4
9 - 35x^2 - 4x^4 = 0
4x^4 + 35x^2 - 9 = 0
(x^2 +9)(4x^2 - 1) = 0
x^2 = -9 or x^2 = 1/4
x = ± 3i or x = ± 1/2
let z = x^-2
then z^2 = x^-4
9 z^2 - 35 z - 4 = 0
(9z+1)(z-4) = 0
z = 4 or z = -1/9
you take it from there
To solve the equation 9x^-4 - 35x^-2 - 4 = 0, we can make use of substitution to simplify it. Let's substitute a variable to simplify the equation:
Let's replace x^-2 with y.
So the equation becomes:
9y^2 - 35y - 4 = 0
Now, we have a quadratic equation which we can easily solve using factoring, completing the square, or using the quadratic formula. In this case, factoring seems to be the most efficient method.
The equation can be factored as:
(3y + 1)(3y - 4) = 0
Now we have two possibilities:
1) 3y + 1 = 0
3y = -1
y = -1/3
2) 3y - 4 = 0
3y = 4
y = 4/3
Since we substituted y for x^-2, we can rewrite these solutions in terms of x:
1) x^-2 = -1/3
Taking the reciprocal of both sides, we get:
x^2 = -3
This equation does not have any real solutions since the square of a real number cannot be negative.
2) x^-2 = 4/3
Taking the reciprocal of both sides, we get:
x^2 = 3/4
Next, we can take the square root of both sides, taking both the positive and negative roots:
x = ±√(3/4)
x = ±√3/2
Hence, the solutions to the original equation 9x^-4 - 35x^-2 - 4 = 0 are:
x = ±√3/2