What value(s) of b will cause 3x2 + bx + 3 = 0 to have one real solution? Select all that apply. b = –6
B. b = –9
C. b = 6
D. b = 9

 -6 and 6

Step-by-step explanation:  

Possible factors that could produce 3x² + 6x +3 are  (x + 1)(3x + 3)

the solution for values of x: x+1=0, x= -1  3x+3=0,  3x= -3 . x= -1 . Same.

Possible factors that could produce 3x² - 6x +3 are  (x - 1)(3x - 3)

the solution for values of x: x-1=0, x= 1  3x-3=0,  3x= 3 . x= 1 . Same.

From the attached graph of the possible equations, you can see that 9 and -9 won't produce ONE real solution. They each have "zeros" at two points on the x-axis. That is two solutions that would make x = 0.

The equations with -6 and +6 each touch the x-axis at only one point each.

im pretty sure its

(8,15)

step-by-step explanation:

answer: the answer is d