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Let a,b be arbitary real numbers. Find the smallest natural number `b` for which the equation x^2+2 (a+b)+(a-b+8)=0 has unequal real roots for all a belongs to real numbers

Deep shikha , 7 Years ago
Grade 11
anser 1 Answers
Dinesh Nayak

Last Activity: 7 Years ago

we have , 
                x2+2(a+b)+(a-b+8)=0
        =>    x2+3a+b+8=0
 
for the equation to have unequal real roots then it should have the discriminant b2-4ac>0 
so we get, 
      b2-4ac>0
=> 0-4(1)(3a+b+8)>0
=> -12a-4b-32>0
 
on solving the in equation we get the range for ‘b’ as
  b \epsilon ( - \infty , -3a-8) 
 
 so according to the range if a = -(8/3)
then b=0
so the smallest natural number for b is 0

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