Thread: Roots of a Sixth-Degree Polynomial

Consider the equation x^6 - 5x^4 + 16x^2 - 72x +9 = 0. I calculate its 2nd derivative 30x^4 - 60x^2 + 32 which is always >0. So f '(x) is monotonic increasing for all x. So f '(x) has only 1 real root. This suggests f(x) has either 0 or 2 real roots; and this is where I am stuck. Since Descarte's rule applied on f(x) says there is 0 or 2 or 4 +ve roots and 0 or 2 -ve roots, it does not help me anyway. How can I go further?

Originally Posted by Sambit

Consider the equation x^6 - 5x^4 + 16x^2 - 72x +9 = 0. I calculate its 2nd derivative 30x^4 - 60x^2 + 32 which is always >0. So f '(x) is monotonic increasing for all x. So f '(x) has only 1 real root. This suggests f(x) has either 0 or 2 real roots; and this is where I am stuck. Since Descarte's rule applied on f(x) says there is 0 or 2 or 4 +ve roots and 0 or 2 -ve roots, it does not help me anyway. How can I go further?

Think Intermediate Value Theorem. (What happens when x=1?)

Intermediate value theorem states that if a function is continuous in an interval [a,b] and if f(a)=/=f(b) then it assumes every value between a and b. Also f(1)<0. Then what? Should I search for values of x such that f(x) becomes +ve?

Originally Posted by Sambit

Intermediate value theorem states that if a function is continuous in an interval [a,b] and if f(a)=/=f(b) then it assumes every value between a and b. Also f(1)<0. Then what? Should I search for values of x such that f(x) becomes +ve?

Yes. (And they shouldn't be hard to find. In fact, you only need one such value, to ensure that the polynomial has at least one root. You already know that the number of roots is either 0 or 2, so that will settle it.)

Yes I got it. Besides, i realised just now that since f ''(x)>0 for all x, the function is convex. So there is either no root or 2 roots. Thank you anyway.