If #a#, #b#, #c#, #d# and #e# are real numbers, prove that the roots of #x^5 + ax^4 + bx^3 + cx^2 + DX + e=0#, cannot be all real if #2a^2<5b#?

1 Answer
May 6, 2018

One of the consequences of Rolle's theorem is that if a differentiable function #f(x)# has real roots at #x=p# and #x=q# (#p< q#), then #f^'(x)# has a real root in #(p,q)#

Thus, if #f(x)# has five real roots, then #f^'(x)# must have at least 4. (In our example, #f^'(x)# is a fourth degree polynomial - and so it must have exactly 4 roots). Continuing, in this fashion #d^3/dx^3f(x)=0# must have two real roots. But for

#f(x) = x^5 + ax^4 + bx^3 + cx^2 + dx + e#

we have

#d^3/dx^3 f(x) = 5times 4 times 3 x^2+a times 4 times 3 times 2x+b times 3 times 2 times 1 #
#qquad qquad = 60x^2+24ax+6b#

So, #d^3/dx^3f(x)=0# becomes

#10x^2+4ax+b=0#

For this to have two real roots, we must have

#(4a)^2>=4 times 10 times b = 40b implies 2a^2 >= 5b#

This is a necessary condition for #x^5 + ax^4 + bx^3 + cx^2 + dx + e = 0# to have five real roots.

Hence if #2a^2<5b#, all five roots of this equation can not be real.