least integer value of a

3x⁴ + 9x³ - 4x³ -12x² +5a = 0
=> (3x³ - 4x²)(x+3) + 5a = 0

=> x²(3x-4)(x+3) = -5a

Consider graph of y= x²(3x-4)(x+3)

find the min. value of dis.

find the least positive value to be added so that the graph just touches x-axis at one / two points.. (hvnt calculated)...and hence calculate 'a' which is the least positive 'integral' value.

the function F(x)= 3x^4 + 4x^3 - 12x^2 ,at x=0,1 and -2 the slope is zero///
the value at x=2 is -32 and it is the minimum value...so the min value of a =7