Question

How do you solve for `x` in the equation `x^4 - 27x^2 + 50 = 0`?

Answer 1

`(x^2 - 25)(x^2 - 2) = 0`

`x^2 - 25 = 0 and`x^2 - 2 =0 #

`x = +-5 and +- sqrt(2)`

Hopefully this helps!

Answer 2

`x = +- sqrt2, and x = +- 5`

Explanation:

This equation is called a bi-quadratic equation.
Call `x^2 = X` and solve the quadratic equation:
`f(X) = X^2 - 27X + 50 = 0.`
Find 2 numbers (real roots) knowing their sum (-b = 27) and their product (c = 50). They are: 2 and 25.
`X = x^2 = 2`--> `x = +-sqrt2`
`X = x^2 = 25` --> `x = +- 5`