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

2 Answers
Noah G
Jul 21, 2016

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

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

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

Hopefully this helps!

Nghi N.
Jul 21, 2016

#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#

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