What are the solutions of the equation x^4 + 3x^2 – 20?

To find the solutions of the equation x⁴ + 3x² – 20 = 0, we first notice that this is a polynomial equation. We can make a substitution to simplify it.

Let y = x². Then, the equation becomes:

y² + 3y – 20 = 0

Now, we can solve this quadratic equation using the quadratic formula:

y = rac{-b ext{ ± } ext{√}(b² – 4ac)}{2a}

In our case, a = 1, b = 3, and c = -20. Plugging these values into the formula gives:

y = rac{-3 ext{ ± } ext{√}(3² – 4 imes 1 imes (-20))}{2 imes 1}

Calculating the discriminant:

3² – 4 imes 1 imes (-20) = 9 + 80 = 89

Now we can find the two possible values of y:

y = rac{-3 ext{ ± } ext{√}(89)}{2}

Next, we simplify these two roots:

y₁ = rac{-3 + ext{√}(89)}{2}

y₂ = rac{-3 – ext{√}(89)}{2}

Since y = x², we need to solve for x by substituting back:

x² = y₁ = rac{-3 + ext{√}(89)}{2}

This gives us:

x = ± ext{√}igg(rac{-3 + ext{√}(89)}{2}igg)

For y₂:

x² = y₂ = rac{-3 – ext{√}(89)}{2}

This will not yield real solutions because the value is negative. Therefore, we only consider the solutions from y₁.

In conclusion, the solutions to the original equation x⁴ + 3x² – 20 = 0 are:

x = ± ext{√}igg(rac{-3 + ext{√}(89)}{2}igg)