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Q.

If x1,  x2 are roots of x2+x+2=0  and  x3,  x4 are roots of x2+3x+4=0 then the value of xi+xj  for  i,  j1,  2,  3,  4  &  i<j  is  λ  then  λ  is 

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answer is 132.

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Detailed Solution

The roots are x1=-1+i72,x2=-1-i72,x3=-3+i72,x4=-3-i72

Consider the expression

x1+x2x1+x3x1+x4x2+x3x2+x4x3+x4

By subsituting the above values,  we get the expression value as 132
 

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