If `2a+3b+6c=0,` then prove that at least one root of the equation `a x^2+b x+c=0` lies in the interval (0,1).

If a+b+c=0 , then, the equation 3ax^(2)+2bx+c=0 has , in the interval (0,1).

Consider the quadratic equation (c-3) x^2 - 2cx + (c - 2) = 0 , c ne 3 . Let S be the set of all integral values of c for which one root of the equation lies in the interval (0, 2) and its other root lies in the interval (2, 3). Then the number of elements in S is:

Consider the quadratic equation (c - 5)x^(2) - 2cx + (c - 4) = 0, c ne 5 . Let S be the set of all integral values of c for which one root of the equation lies in the interval (0, 2) and its other root lies in the interval (2, 3). Then the number of elements in S is

If exactly one root of the quadratic equation x ^(2) -(a+1 ) +2a =0 lies in the interval (0,3) then the set of value 'a' is given by :

If a+b+c=0, a,b,c in Q then roots of the equation (b+c-a) x ^(2) + (c+a-c) =0 are:

If a+b + c = 0, then show that the quadratic equation 3ax^(2) + 2bx +c=0 has at least one root in [0,1].

Find the condition if the equation 3x^(2)+4ax+b=0 has at least one root in (0,1).