How do you find all the real and complex roots of #x^4-6x^2+5=0#?

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Answer 1 · 2016-01-22T16:32:58

#x=-sqrt5,-1,1,sqrt5#

Notice that this is very similar to #x^2-6x+5#, which could be factored into #(x-5)(x-1)#. Similarly, this polynomial can be factored as

#(x^2-5)(x^2-1)=0#

To solve this, set each individual part being multiplied equal to #0# and solve.

#x^2-5=0#
#x^2=5#
#x=+-sqrt5#

#x^2-1=0#
#x^2=1#
#x=+-1#

These are the equation's four roots (where the equation crosses the #x# axis). We can check a graph:

graph{x^4-6x^2+5 [-19.53, 21.01, -6, 14.28]}

Note that #sqrt5approx2.24#.