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Grade: 12
        X*4+ax*3++bx*2+cx+d be a polynomial equation with real cofficients and real zeroes.if |f(i)|=1.find a+b+c+d.i=root of -1
6 years ago

Answers : (3)

Sher Mohammad
IIT Delhi
askIITians Faculty
174 Points
							f(i)=1-ai-b-ci+d=1
1-b-d=real part=1
-i(a+c)=complex part =0
b+d=0
a+c=0
a+b+c+d=0
sher mohammad
b.tech, iit delhi
6 years ago
mycroft holmes
272 Points
							If p,q,r,s are the real roots of the polynomial, then f(i) = (i-p)(i-q)(i-r)(i-s)
and hence |f(i)| = (1+p^2)(1+q^2)(1+r^2)(1+s^2) >=1.

But |f(i)| = 1 and hence we must have p=q=r=s=0
i.e. a=b=c=d=0=a+b+c+d
						
6 years ago
AMRENDRA JHA
37 Points
							@sher md. its mod of f(i).
						
6 years ago
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