IIT JEE , AIEEE Forum - Mathematics , Algebra

hsbhatt

f(z) is a non-constant polynomial with real coefficients such that for some real number a, we have

$f(a) \ne 0, f$

Prove that f(z) cannot have all roots real.

I happen to know the official solution. So, only original looking solutions get rated.

feynmann

write $f(x)=c\prod(x-\alpha_{i})}$ ,where $\alpha_{i}$ s are the roots of the polynomial

given that f(a) !=0 whch means none of the roots are a .

Now we get f'(a)=0

so taking logarithm of both sides and then taking derivative at x=a , we obtain

$\sum{\frac{1}{a-\alpha_{i}}=0$ ......(2)

again taking the derivative of f'(x ) at x=a and imposing the condition that f''(a)=0 , we get using (2)

$\sum{\frac{1}{(a-\alpha_{i})^2}=0$ ....(3)

but it is given that a is real . So if all the roots are real then surely (3)can't hold true .

Hence proved .

hsbhatt

thats the solution i was looking for!

A bit of clarification may be required here:

We have for a polynomial f(x) with roots $\alpha_i$ :

$\frac{f$

Since f'(a) = f"(a) = 0, we have

$\sum_{i=1}^n \frac{1}{(x-\alpha_i)^2} = 0$