Help!

Using the distance formula...

the distance between  (0, 0)  and  ( 3√5, d+3 )   \(=\,\sqrt{(3\sqrt5-0)^2+(d+3-0)^2} \\~\\ =\,\sqrt{(3\sqrt5)^2+(d+3)^2} \\~\\ =\,\sqrt{45+d^2+6d+9} \\~\\ =\,\sqrt{d^2+6d+54}\)

And they tell us that this equals  3d  ....

\(\sqrt{d^2+6d+54}\,=\,3d \\~\\ d^2+6d+54\,=\,9d^2 \\~\\ 0\,=\,8d^2-6d-54 \\~\\ 0\,=\,8d^2-24d+18d-54 \\~\\ 0\,=\,8d(d-3)+18(d-3) \\~\\ 0\,=\,(d-3)(8d+18) \\~\\ d=3\qquad\text{or}\qquad d=-\frac94\)       Now we just need to solve this equation for  d  .

But... -9/4 is extraneous...so....the smallest possible value of  d  is  3 .

Edited to fix my error....thanks CPhill

( 3√5, d + 3)  =   (√45 , d + 3 )

So we have that

3d  = √  [  (√ 45)^2  +  (d + 3)^2 ]

3d  =  √ [  45  +  d^2 + 6d + 9 ]

3d  =  √ [ 54 + d^2 + 6d ]       square both sides

9d^2   = d^2 + 6d + 54      rearrange as

8d^2 - 6d - 54  = 0

4d^2  - 3d  -  27  =  0      factor

(4d + 9) (d - 3)  = 0

Setting both factors to 0  and solving for d, we have that

d  = -9/4     or  d  =  3

The first value  gives the  distance  of 3(-9/4)  =  -27/4 units.....but distance is  a positive quantity

So......d  = 3   is the correct value for d  and  3d  =  9 units