Value of a for which sum of squares of roots of equation x^2-(a-2)x-a-1=0 assumes least value.
A)0
B)1
C)2
D)3

• a=1
  
  A quadratic equation is of the form,
  x^2 - (sum of roots)x + (product of roots) = 0
  
  Here,
  Sum of roots = a-2
  product of roots = -a-1
  
  Let the roots of the equation be p and q
  p+q = a-2
  p*q = -a-1
  
  (p+q)^2 = (a-2)^2
  p^2 + q^2 + 2pq = a^2 + 4 -4a
  p^2 + q^2 = a^2 + 4 - 4a - 2pq
  
  sum of squares of roots = a^2 + 4 -4a - 2(-a-1)
  =a^2 +6 -2a
  
  d/dt(a^2 + 6 -2a) = 0 [at minimum value of a]
  2a - 2 = 0
  a = 1
  
• 10 years agoHelpfull: Yes(0) No(1)
• a=1
  
  let x & y are roots of the quadratic equation x^2-(a-2)x-a-1=0
  then we say that, x + y = -(-(a-2))/1 = a-2
  and x * y = -(a+1)/1=-(a+1);
  
  GIVEN, x^2 + y ^2 has least value
  so, x^2 + y^2=(x+y)^2-2*x*y
  x^2 + y^2=(a-2)^2-2*(-(a+1))=a^2-2a+6
  
  FOR minimum value of x^2 + y^2
  
  d(a^2-2a+6)/d(a)= 2a-2
  then d^2(2a-2)/d(a)=2 > 0
  hence we can say that, at 2a-2 = 0 ,we get minmum value of a.
  so, 2a-2=0
  a=1
  
  
• 10 years agoHelpfull: Yes(0) No(0)