How do you find all the real and complex roots of #x^4-14x^2 +45=0#?

1 Answer
Mar 3, 2016

Answer:

Equation has all real roots given by #x=-{sqrt5, sqrt5, -3, 3}#

Explanation:

Let us first factorize #x^4−14x^2+45#, which can be done by splitting middle term into #-9x^2# and #-5x^2#. Hence,

#x^4−14x^2+45=0#

#hArr# #x^4−9x^2-5x^2+45=0#

#hArr# #x^2(x^2−9)-5(x^2-9)=0#

#hArr# #(x^2-5)(x^2−9)=0# and using #a^2-b^2=(a+b)(a-b)#, this becomes

#hArr# #(x+sqrt5)(x-sqrt5)(x+3)(x-3)=0#

or #x=-{sqrt5, sqrt5, -3, 3}# and hence equation has all real roots.