Welcome Visitor! Login | Sign Up | Help!
Refer a Friend
|
Follow us on:
More
Home » Ask & Discuss » Mathematics. » Algebra « Back to Discussion
Algebra
f(z) is a non-constant polynomial with real coefficients such that for some real number a, we have
Prove that f(z) cannot have all roots real.
I happen to know the official solution. So, only original looking solutions get rated.
write ,where 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
......(2)
again taking the derivative of f'(x ) at x=a and imposing the condition that f''(a)=0 , we get using (2)
....(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 .
thats the solution i was looking for!
A bit of clarification may be required here:
We have for a polynomial f(x) with roots :
Since f'(a) = f"(a) = 0, we have
Name
Mobile No.
Physics.
General Physics
Mechanics
Optics
Thermal Physics
Electricity
Magnetism
Modern Physics
Mathematics.
Trignometry
Analytical Geometry
Differential Calculus
Integral Calculus
Vectors
Chemistry.
Physical Chemistry
Organic Chemistry
Inorganic Chemistry
Biology
Botany
Zoology
Institutes
Coaching Institutes & Course Material
About IITs and JEE
Non IIT Institutes
Counselling Zone
Parents
Parent Discussion Board
Board
Course Material
Computer Science
Fun Zone
Lounge
Games, Puzzles and Quizzes
General Knowledge
write
,where
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
again taking the derivative of f'(x ) at x=a and imposing the condition that f''(a)=0 , we get using (2)
but it is given that a is real . So if all the roots are real then surely (3)can't hold true .
Hence proved .