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Question

Select the values of s  for which polynomial p(x)=sx2+x(s1)27  satifies p(1)=0
  1. 1
  2. 3
  3. -2
  4. 2


Solution

The correct options are
B 3
C -2
Given p(x)=sx2+x(s1)27  &  p(1)=0p(1)=s×12+1×(s1)27=0s+(s1)27=0s+s22s+17=0s2s6=0

Lets solve for s.
s23s+2s6=0s(s3)+2(s3)=0(s3)(s+2)=0s=3,2

So, for s=3,2 we get p(1)=0

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