If the graph of the quadratic function f(x) = x^2 + dx +3d has its vertex on the x axis what are the possible values of d? what if f(x) = x^2 + 3dx +1?

The coordinates of vertex are: ( x , 0 ) and x = -b/2a
a ) For the 1st function: x = -d/2
0 = d² / 4 - d² / 2 + 3 d   / · 4
0 = - d² + 12 d
d ( 12 - d ) = 0,    d = 0   or d = 12
If d = 0,  f (x) = x²
If d 0 12  f (x) = x² + 12 x + 36 = ( x + 6 )²
b ) For the 2nd function:
x = - b/2a = - 3 d/2
0 =  9 d² / 4  - 9 d / 2 + 1   /  · 4
0  = 9 d²  - 18 d + 4

d = 1 - √13/2  or d  = 1 + √13/2

you trying to figure out what is B?

B is isosceles triangles