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.