Question

The maximum possible number of real roots of equation

x⁵ – 6x² – 4x + 5 = 0 is -

• A. 0
• B. 3
• C. 4
• D. 5

Answer 1

Answer: (B)

3

Answer 2

f(x) = x⁵ – 6x² – 4x + 5

=

Number of change of sign = 2

Atmost 2 positive real roots

f(–x) = – x⁵ – 6x² + 4x + 5

$- \overbrace { + }$

Number of change of sign = 1

Atmost 1 negative real root

or

+ ve real root (max.)	– ve real root (max.)	Imaginary root
2	1	2

Þ (Max.) real roots = 3