How do you find the roots of #x^4-5x^2+4=0#?

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Answer 1 · 2018-03-22T04:45:05

#x = +- 2, +-1#

Let #y = x^2#.

#y^2 - 5y + 4 =0#

This is a nice and simple trinomial to factor.

#(y - 4)(y - 1) =0 #

#y = 4 or y = 1#

We now reverse our substitution.

#x^2 = 4 or x^2 =1#

#x = +- 2 or +-1#

The graph of the function #f(x) = x^4 - 5x^2 + 4# confirms that there are zeroes at #x = +-2 and +-1#.

graph{x^4 - 5x^2 + 4 [-10, 10, -5, 5]}

Hopefully this helps!