-3x^4-3x^2+1

Question

What are the roots of the function?

Answer 1

Use substitution method.

 

Let        q = x²

           q² = x⁴

       

Then,

 

-3q² - 3q + 1 = 0

 

 

Divide both sides of the equation by -1 to hopefully factor using FOIL.

 

3q² + 3q - 1 = 0

 

 

Using the quadratic formula to solve for q,

 

q = (-3 ± √(9 - 4(-3))) / 6

 

q = (-3 ± √(21)) / 6

 

q = (-3 - √21) / 6               and              q = (-3 + √21) / 6

 

 

Then substitute these values of q into the original substitution.

 

x = ±√q

 

 

x₁ = - √[(-3 - √21) / 6]

 

x₂ = √[(-3 - √21) / 6]

 

x₃ = - √[(-3 + √21) / 6]

 

x₄ = √[(-3 + √21) / 6]

 

 

The only real solutions are  x₃ and x₄.

Answer 2

This is what's called a quadratic in the variable x².

 

By the quadratic equation,

x² = (-3 ± √21)/6

 

So x = ± the square root of that expression