Question

# The maximum possible number of real roots of equation x5−6x2−4x+λ2=0is

Solution

## f(x)=x5−6x2−4x+λ2=0 +  −  −  + 2 changes of sign ⇒ maximum two positive roots. f(−x)=−x5−6x2+4x+λ2 −       −       +       + 1 change of sign ⇒ maximum one negative roots. ⇒ total maximum possible number of real roots = 2 + 1 = 3.

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