Polynomials .... (1 Viewer)

[ashokkumar]

Suppose f(x) = x^3 + bx + c, where b and c are constants. Suppose that the equation f(x) = 0 has three distinct roots p,q,r.

i) Find p + q + r

ii) find p^2 + q^2 + r^2

iii) Since the roots are real and distinct, the graph of y= f(x) has two turning points, at x = u and x = v, and f(u).f(v) < 0. Show that 27c^2 + 4b^3 < 0.


Don't worry about parts i) and ii). I only need help with part iii)

 

[omgmynameislong]

iii)

If the roots are distinct then it passes through x axis 3 times. Therefore f(u) must be a negative value and f(v) must be a positive value. A positive value times a negative value is alwyas lesser than 0.

therefore f(u).f(v)<0

f'(x) = 3x^2 +b = 0

therefore x= +(-b/3)^1/2 and -(-b/3)^1/2

so u=+(-b/3)^1/2 v=-(-b/3)^1/2

f(u).f(v) <0

sub in the u and v and you should get that thing youre suppose to show, i cbf.