What value(s) of b will cause 27x^2 +bx+3=0 to have one real solution?

Question

What value(s)? Please help

Answer 1

for your factors you need 2 numbers which multiply to 81 (27 x3) that are the same which would be 9 so b has to be 9+9=18

If you graph y=27x^2 +18x+3 you will see it has only one solution x= -1/3