Complex Numbers And Quadratic Equations question 33
Question: The maximum possible number of real roots of equation $ x^{5}-6x^{2}-4x+5=0 $ is [EAMCET 2002]
Options:
0
3
4
5
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ f(x)=x^{5}-6x^{2}-4x+5=0 $ Then the number of change of sign in $ f(x) $ is 2 therefore $ f(x) $ can have at most two positive real roots. Now, $ f(-x)=-x^{5}-6x^{4}+4x+5=0 $ Then the number of change of sign is 2. Hence $ f(x) $ can have at most two negative real roots. So that total possible number of real roots is 4.
BETA
